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WHITENS   SERIES   OF  MATHEMATICS 


ELEMENTS   OF   GEOMETRY 


PLANE  AND    SOLID 


BY 

JOHN    MACNIE,    A.M. 

Author  of  "  Theory  of  Equations  " 


EDITED  BY 

EMERSON   E.   WHITE,  A.M.,  LL.D. 

Author  of  "White's  Series  of  Mathematics 


NEW   YORK:- CINCINNATI:.  CHICAGO 

AMERICAN    BOOK    COMPANY 


1^1 ^ 

White's  Series  of  Mathematics. 

ORAL  LESSONS  IN  NUMBER.     (For  Teachers.) 


FIRST  BOOK  OF  ARITHMETIC. 
NEW  COMPLETE  ARITHMETIC. 
SCHOOL  ALGEBRA.     {In  Preparation.) 
ELEMENTS    OF  GEOMETRY. 

ELEMENTS  OF  TRIGONOMETRY.   (//»  Preparation.) 
* 


COPYEIGHT,   1895,   BY   AMERICAN   BoOK  COMPANY. 


|>rinteb  at 
TTbe  Eclectic  l^reee 
(Cincinnati.  TIL  S.  a 


PREFACE. 


In  this  treatise,  an  endeavor  is  made  to  present  the  elements  of 
geometry  with  a  logical  strictness  approaching  that  of  Euclid,  while 
taking  advantage  of  such  improvements  in  arrangement  and  nota- 
tion as  are  suggested  by  modern  experience.  It  has  been  carefully 
kept  in  mind  that  the  purpose  of  such  a  work  is  only  in  a  secondary 
degree  the  presentation  of  a  system  of  useful  knowledge.  A  much 
more  important  purpose  is  to  afford  those  who  study  this  subject 
the  only  course  of  strict  reasoning  with  which  the  great  majority 
of  them  will  ever  become  closely  acquainted.  A  mind  that,  by  exer- 
cise in  following  and  weighing  examples  of  strict  logical  deduction, 
has  learned  to  appreciate  sound  reasoning,  and,  by  practice  on 
suitable  exercises,  has  been  trained  to  reason  out  a  sound  logical 
deduction  for  itself,  has  gained  what  is  of  far  greater  importance 
than  mere  knowledge ;  it  has  gained  power.  A  treatise  on  rational 
geometry  ought,  accordingly,  to  have  for  guiding  principles  those 
laid  down  by  Pascal  as  the  chief  laws  of  demonstration,  substan- 
tially as  follows :  to  leave  no  obscure  terms  undefined ;  to  assume 
nothing  not  perfectly  evident ;  to  prove  everything  at  all  doubtful, 
by  reference  to  admitted  principles. 

In  accordance  with  the  first  principle,  great  care  has  been  taken 
in  the  wording  of  the  definitions.  In  the  case  of  some  terms,  such 
as  straight  line  and  angle,  for  which  no  definitions  quite  free  from 
objection  have  as  yet  been  proposed,  those  adopted  have  been 
chosen,  not  as  theoretically  perfect,  but  as  best  suited  to  the  com- 
prehension of  the  beginner,  and  most  available  in  deducing  the 
properties  of  the  things  defined. 

The  use  of  hypothetical  constructions  has  been  abandoned  for 
several  reasons.  To  assume  them  silently,  as  is  now  usually  done, 
is  unwarrantable  in  a  treatise  upon  a  science  supposed,  above  all 
others,  to  consist  of  a  series  of  rigorous  deductions  from  admitted 
truths.    Why  state  so  carefully  that  we  must  assume  the  possibility 

3 


183969 


4  PREFACE. 

of  drawing  or  prolonging  a  straight  line,  and  say  nothing  in  regard 
to  constructions  so  much  less  obvious?  The  author's  experience 
in  teaching  geometry  has  convinced  him  that  these  stealthy  assump- 
tions are  decidedly  adverse  to  the  acquisition,  on  the  part  of  the 
learner,  of  those  habits  of  strict  reasoning  which  it  is  the  main 
object  of  geometrical  study  to  impart.  The  teacher  should  have 
it  in  his  power  to  inquire,  at  every  step,  not  only  why  such  a  state- 
ment is  true,  but  also  why  such  a  construction  is  allowable.  On 
the  other  hand,  the  insertion  of  the  few  problems  really  needed 
as  auxiliaries  in  the  demonstration  of  theorems,  imparts  to  their 
sequence  a  logical  consistency  that  cannot  obtain  where  the  learner 
is  continually  required  to  perform  operations,  the  possibility  of 
which  has  neither  been  proved  nor  formally  assumed. 

In  regard  to  the  use  of  circumferences  as  construction  lines  in 
the  first  book,  it  may  be  remarked  that,  of  all  lines,  the  circum- 
ference is  the  easiest  to  define,  to  construct,  and  to  conceive,  —  far 
more  so  than  is  the  case  with  the  straight  line.  No  property  of 
the  circle,  not  even  its  name,  is  introduced  in  the  first  book,  but 
only  that  property  of  the  circumference  which  is  given  in  its 
definition,  and  some  immediate  consequences  from  that  property. 
The  employment,  at  the  outset,  of  the  simpler  of  the  two  lines 
treated  of  in  elementary  geometry  would  hardly  require  apology 
or  explanation  but  for  the  force  of  custom.  Yet,  strangely  enough, 
in  treatises  that  scrupulously  defer  the  definition  of  the  terms  cir- 
cumference and  radius  to  a  subsequent  book,  there  seems  to  be  no 
scruple  against  the  employment  of  arcs  in  illustration  of  the  nature 
of  angles,  and  we  see  gravely  laid  down  the  postulate :  A  circum- 
ference may  be  described  about  any  point  as  center,  etc. 

The  deviation  in  this  work  from  the  usual  order  of  propositions 
is  comparatively  slight.  In  the  early  part  of  the  first  book,  so 
important  as  the  foundation  of  the  science,  the  properties  of  tri- 
angles are  introduced  immediately  after  the  discussion  of  the  general 
properties  of  angles.  This  arrangement,  especially  as  regards  the 
different  cases  of  equal  triangles,  presents  several  advantages  :  these 
propositions  are  immediately  deducible  from  first  principles,  or  from 
each  other ;  they  are  easily  grasped  by  the  beginner ;  above  all,  they 
are  of  the  highest  utility  as  aids  to  further  acquisition.  It  is  in 
itself  no  slight  advantage  for  the  learner  to  become  accustomed, 
from  the  first,  to  the  use  of  these  important  auxiliaries  in  demon- 


PREFACE.  5 

stration.  From  the  second  book,  again,  certain  propositions  that 
treat  of  proportional  angles  have  been  removed  to  the  place  where 
they  belong,  after  the  discussion  of  ratio  and  proportion.  In  the 
treatment  of  these  subjects,  while  adhering  to  the  now  prevalent 
method,  an  endeavor  has  been  made  to  obviate  one  frequent  source 
of  confusion  by  making  a  clear  distinction  between  concrete  quan- 
tities and  their  numerical  measures. 

In  regard  to  propositions  and  corollaries,  the  rule  observed  has 
been  to  admit  only  such  as  are  important  in  themselves,  or  have 
a  bearing  on  subsequent  demonstrations  and  studies.  In  this 
era  of  over-crowded  curricula,  the  aim  of  an  elementary  text-book 
should  be  to  present  the  necessary  rather  than  the  novel  or  merely 
interesting.  For  this  reason  some  subjects  have  been  relegated  to 
an  appendix,  where  they  may  be  studied  or  omitted  according  to 
circumstances.* 

The  exercises  have  been  carefully  selected  with  a  view  to  their 
bearing  upon  important  principles,  and  are,  with  few  exceptions, 
of  such  slight  difficulty  as  not  to  discourage  the  learner  of  average 
ability.  In  the  first  sets  of  exercises,  ample  assistance  is  afforded 
the  student  by  means  of  references  and  diagrams,  —  aids  that  are 
withheld  from  the  point  where  the  student  should  have  learned  to 
help  himself.  It  is  by  no  means  expected  that  the  average  class  will 
find  time  for  all  the  exercises,  but  enough  are  given  to  afford  the 
teacher  full  opportunity  of  choice. 

Suggestions.  It  is  earnestly  recommended  that,  before  any  book 
work  is  assigned  to  a  beginning  class,  a  lesson  be  devoted  to  the 
constructions  given  in  Arts.  200-206.  The  teacher,  having  shown 
on  the  blackboard,  for  example,  how  to  bisect  a  straight  line, 
should  set  the  class  to  doing  the  same,  and  require  each  pupil  to 
bring  in,  next  day,  one  or  more  neatly  worked  examples  of  the 
required  constructions.  The  practical  familiarity  thus  gained  with 
the  geometrical  concepts  involved  will  amply  repay  the  time  thus 
spent.  The  two  great  sources  of  difficulty  to  the  beginner  in 
geometry  are  the  comparative  novelty  of  the  subject  matter  and 
the  unaccustomed  clearness  of  conception  and  exactness  of  expres- 


*  The  whole  of  the  last  part  of  Book  IX.,  treating  of  spherical  angles 
and  polygons,  may  as  well  be  omitted  by  pupils  not  to  take  up  spherical 
trigonometry. 


6  PREFACE. 

sion  required  in  this  new  study ;  it  will  be  found  that  the  second 
source  of  difficulty  is  most  easily  diminished  by  reducing  the  first 
to  a  minimum. 

The  easy  exercises  at  the  foot  of  the  page  can  be  employed  to 
the  best  advantage  as  material  for  impromptu  work,  the  teacher 
giving  as  much  aid,  by  suggestive  questions,  as  will  enable  the 
class  to  solve  them  oifhand.  Such  as  prove  too  difficult  for  this 
first  attempt  may  then  be  assigned  as  work  to  be  prepared.  The 
questions  found  at  the  end  of  most  books  may  either  be  taken  up 
as  the  class  progresses  through  the  book,  or  be  left  for  review. 
Such  matter  as  that  found  on  p.  105  is  to  be  carefully  read  over 
in  the  class  with  a  running  comment  by  the  teacher ;  the  gist  of 
the  ideas,  not  the  words  in  which  they  are  expressed,  being  what 
the  pupil  should  try  to  retain. 

The  pupil  should  be  required  to  give  the  exact  words  of  defini- 
tions, axioms,  postulates,  and  enunciations.  In  other  matters  some 
latitude  may  be  allowed,  though  occasion  should  be  taken  to  point 
out  in  what  respects  the  pupil's  own  wording  may  be  objectionable. 

Outside  of  Euclid,  it  is  of  doubtful  utility  to  exact  from  pupils 
a  knowledge  of  all  propositions  by  number.  The  axioms,  postu- 
lates, and  certain  important  propositions  should  be  known  by 
number,  but  in  written  or  oral  work,  other  references  may  be 
given  by  abbreviated  quotations. 


CONTENTS. 

PAET  I.     PLANE   GEOMETEY. 
INTRODUCTION.    DEFINITIONS. 

PAGE 

Lines 13 

Angles 15 

Propositions 18 

Axioms 19 

Postulates .19 

Method  of  Proof 20 

BOOK  I.     LINES  AND  RECTILINEAR  FIGURES. 

Angles 23 

Triangles 28 

Perpendiculars 46 

Parallels 51 

Quadrilaterals 61 

Exercises 69 

Geometrical  Synthesis  and  Analysis 71 

Exercises 74 

BOOK  II.     THE  CIRCLE.    LOCI.    PROBLEMS. 

Elementary  Properties 78 

Chords 81 

Tangent  and  Secants 90 

Exercises 93 

Constructions     ..........  95 

Plane  Loci 102 

Exercises      .                103 

Analysis  of  Problems       ........  105 

Exercises 107 

7 


8  CONTENTS. 

BOOK  III.     RATIO.     PROPORTION.     LIMITS. 

PAGE 

Measurement Ill 

Ratio 113 

Proportion.     Definitions          .......  114 

Exercises 116 

Limits 117 

Theorems  in  Proportion    .        . 119 

BOOK  IV.     PROPORTIONAL  ANGLES  AND   LINES. 
SIMILAR   POLYGONS. 

Proportional  Angles 128 

Exercises 136 

Proportional  Lines 138 

Similar  Polygons 144 

Ratios  of  Certain  Lines 151 

Exercises 155 

Constructions 157 

Exercises 160 

BOOK  V.     AREAS  AND   THEIR   COMPARISON. 

Quadrilaterals 165 

Exercises .        .        .         .  186 

Constructions .        .        .        .  187 

Numerical  Applications 191 

Exercises 194 

BOOK  VI.     REGULAR  PLANE  FIGURES. 

Regular  Polygons 198 

Area  op  the  Circle 208 

Division  of  Circumference 212 

Exercises 221 


CONTENTS.  9 

PART   II.     SOLID   GEOMETRY. 
BOOK  VII.      PLANES  AND  POLYHEDKAL  ANGLES. 

PAGE 

Planes  and  Perpendiculars     .        .         .         .         .         .         .  225 

Planes  and  Parallels 234 

Dihedral  Angles       .........  240 

Polyhedral  Angles .        .  261 

Exercises .  258 

BOOK  VIII.      POLYHEDRONS. 

Prisms 262 

Pyramids      .         .         .         .         .         ...         .         .        .  275 

The  Regular  Polyhedrons 286 

Exercises 290 

BOOK  IX.  THE  THREE  ROUND  BODIES. 

Cylinders 293 

Cones 297 

Spheres .  300 

Spherical  Angles  and  Polygons      ......  309 

Polar  Triangles 313 

Exercises     . 330 

BOOK  X.  MEASUREMENT  OF  THE  THREE  ROUND  BODIES. 

Cylinders. 332 

Cones 336 

Exercises 342 

Spheres 344 

Exercises 352 

APPENDIX. 

Symmetry 355 

Symmetrical  Polyhedrons 360 

Maxima  and  Minima 362 

INDEX 371 


SYMBOLS. 


•••  because  or  since. 

.'.  therefore. 
+  plus. 
—  minus. 
=  is  equal  to. 
=  is  equivalent  to. 
>  is  greater  than. 
<  is  less  than. 
^  coincides  with. 
II  parallel.* 
±  perpendicular.* 
O  circle. 


Z  angle. 
St.  Z  straight  angle. 
rt.  Z  right  angle. 

A  triangle. 
rt.  A  right  triangle. 
AB,  line  or  side  ab. 
comp.  complement. 
supp.  supplement. 
resp.  respectively. 
Req.  required. 
Q.E.D.  as  teas  ?o  be  proved. 
Q.E.F.'  as  was  to  be  done. 


The  symbols  marked  above  with  an  asterisk  are  to  be  read  as 
nouns  when  preceded  by  an  article  or  similar  word.  Plurals  are 
formed  by  adding  's  to  the  singular.    Thus,  A's,  ll's,  etc. 

In  reading  such  expressions  as  A  A'B'C,  O  abed,  it  will  be  suffi- 
cient, in  almost  every  case,  to  give  the  distinguishing  epithet  to  the 
last  letter  only.  Thus,  the  above  expressions  may  be  read  respec- 
tively :  triangle  ABC  prime,  circle  abed  small  or  minor.  It  is,  of 
course,  seldom  necessary  to  say  the  line  or  the  side  AB ;  it  is  usually 
sufficient  to  say  AB. 


10 


PART   I.     PLANE   GEOMETRY. 


>i*;c 


INTRODUCTION. 


// 


We  conceive  space  as  extending  without  limit  in  every 
direction.  Each  physical  body,  a  marble  cube,  for  example, 
occupies  a  limited  portion  of 
space.  Thus  the  cube  AG  is 
limited  on  all  sides  by  bound- 
aries which  are  called  faces;  as 
AFj  EG,  etc.  These  faces,  again, 
meet  in  edges,  which  are  called 
lines;  as  ^5,  i^G^,  etc. ;  and  final- 
ly, these  lines  meet  in  extremi- 
ties, which  are  called  points; 
as  A,  F,  H,  etc.  If,  now,  leav- 
ing its  material  entirely  out  of  consideration,  we  think  only 
of  the  portion  of  space  occupied  by  the  cube,  with  its  faces, 
edges,  and  points,  we  have  before  us  a  so-called  geometrical 
solid.  It  is  with  such  abstract  solids  as  this  that  geometry 
is  concerned. 


/ 

/ 

/ 

F 

V 

DEFINITIONS. 

1.  A  solid  is  a  limited  portion  of  space,  and  has  three 
dimensions  :  length,  breadth,  and  thickness. 

2.  Surfaces  are  the  limiting  boundaries  of  solids,  and 
have  two  dimensions  :  length  and  breadth. 

n 


12  PLANE   GEOMETRY. 

3.  Lines  are  the  boundaries  or  intersections  of  surfaces, 
and  have  but  one  dimension :  length. 

4.  Points  are  the  extremities  or  intersections  of  lines ; 
hence  a  point  has  no  dimension,  but  only  position. 

Just  as  we  conceive  a  geometrical  solid  apart  from  mate- 
rial, so  we  may  conceive  of  surfaces  apart  from  solids,  of 
lines  apart  from  surfaces,  and  of  points  apart  from  lines. 
These  abstract  points,  lines,  surfaces,  and  solids  are  the 
so-called  space-concepts,  the  elements  of  our  notions  of  space. 

5.  Geometry  is  the  science  that  treats  of  the  proper- 
ties and  relations  of  space-concepts,  also  called  geometrical 
concepts. 

Each  geometrical  concept  has  position,  determined  by  the 
location  of  its  points  in  space ;  and  all  except  points  have 
form  or  shape,  determined  by  the  relative  position  of  their 
points,  and  extent  or  magnitude,  determined  by  the  nearness 
or  remoteness  of  their  bounding  points. 

6.  A  straight  line  is  a  line  that  does  not  change  its  direc- 
tion at  any  point ;  as  AB. 


7.  A  Iroken  line,  as  J  5  (7  or  ^  J5  CDE, 
changes  direction  at  one  or  more  points. 

8.  A  curved  line,  as  MN,  changes 
direction  at  every  point ;  i.e.,  no  part  of  it  is  a  straight  line. 
The  curve  is  said  to  be  closed  when  it  forms  a  continuous 
boundary.      Thus  P  and  Q  are  closed  curves. 


CZ3 


The  term  line,  when  used  hereafter  by  itself,  is  to  be 
understood  as  denoting  a  straight  line;  similarly,  curve  will 
signify  curved  line. 


INTRODUCTION.  13 

9.  A  plane  surface  or  plane  is  such  that  a  straight 
line  passing  through  any  two  points  in  that  surface  lies 
wholly  in  the  surface. 

10.  A  geometrical  figure  is  any  given  combination  of 
points,  lines,  or  surfaces ;  the  representation  of  a  figure  is 
a  diagram. 

Just  as  a  diagram  is  the  representation  of  a  figure,  so  the 
lines  of  the  diagram  and  the  surface  on  which  it  is  drawn 
are  merely  more  or  less  imperfect  representations  of  the 
ideally  perfect  lines  and  planes  to  be  treated  of. 

11.  K  plane  figure  is  such  that  all  its  points  lie  in  the 
same  plane. 

12.  Plane  geometry  is  the  geometry  of  plane  figures. 

13.  A  magnitude  is  a  concept  any  part  of  which  is  of  the 
same  kind  as  the  whole. 

Thus  since  any  part  of  a  line  is  a  line ;  any  part  of  a 
surface,  a  surface  ;  any  part  of  a  solid,  a  solid,  —  lines,  sur- 
faces, solids,  and  angles,  presently  to  be  defined,  are 
geometrical  magnitudes.  Other  magnitudes  are  time,  weight, 
mass,  etc. 

14.  Equal  magnitudes  are  such  as  can  be  made  to  coin- 
cide exactly. 

Magnitudes  coincide  exactly  when,  one  being  placed  upon 
the  other,  every  point  of  the  one  lies  upon  a  corresponding 
point  of  the  other. 

LINES. 
From  the  definition  of  the  straight  line  {fo),*  it  follows  :  — 

15.  Straight  lines  that  coincide  in  part  coincide  throughout. 

16.  Two  points  determine  a  straight  line. 

17.  Two  straight  lines  cannot  inclose  a  surface. 

*  Numbers  in  parentheses  throughout  the  book  refer  in  general  to 
paragraphs. 


14 


PLANE   GEOMETRY. 


For  if  two  lines  ABC,  ABD,  coincide  in  a  part  AB  only, 
or  in  two  points  A  and  B  only,  they  cannot  both  be  straight 
lines  (6),  since  one,  at  least,  must  change  direction. 


18.  Since  the  direction  of  one  point  from  another  may 
be  regarded  as  the  path  of  a 

point   that    passes    along    the   ^ ^ 

straight  line  that  connects  these 

points,  and  as,  if  A  and  B  are  the  points,  the  direction  may 
evidently  be  either  from  A  to  B  or  from  B  to  A,  the  same 
straight  line  may  mark  either  of  two  opposite  directions, 
or  it  may  be  regarded  as  extending  in  both  directions  from 
any  point  in  it. 

19.  A  line  is  said  to  be  indefinitely  produced  when  pro- 
longed as  far  as  necessary,  or  without  assigned  limit.  If 
two  straight  lines  AB,  CD,  lie  in  the  same  plane,  it  is 
evident  that,  if  indefinitely  produced,  they  must  either  meet 
or  not  meet.     Thus  :  — 

(a)  If  they  can  be  produced  so  as  to  have  more  than  one 
common  point,  they  lie  in 

the  same  line   and  have     ^ 

the  same  directions. 

(6)  If  they  can  have 
hut  one  common  point, 
they  can  intersect  in  that 
point,  and  they  have  dif- 
ferent directions. 

A B 

(c)  If   they   can    have 
no  common  point,  —  that  is,  cannot  meet,  —  they  are  said  to 
be  parallel. 


INTRODUCTION. 


15 


20.  If  the  straight  lines  drawn  from  A  io  B  and  from  z> 
to  E  are  equal,  the  distance  from  A  to 
B  is  said  to  be  equal  to  that  from  B  to 
E.  If  the  distances  from  ^  to  ^  and  C 
are  equal,  A  is  said  to  be  equidistant 
from  B  and  C,  while  5  and  C  are  said 
to  be  equally  distant  from  ^. 

21.  A  circumference  is  a  closed  curve  described  in  a  plane, 
and  is  such  that  all  its  points  are  equally  distant  from  a 
point  within  the  curve,  called  the  center. 

22.  A  radius  is  any  straight  line  drawn 
from  center  to  circumference.  Thus  OA, 
OB,  OC,  are  radii  of  the  circumference 
ACB,  whose  center  is  0. 

23.  All  radii  to  the  same  circumference 
are  equal.  It  is  also  obvious  that  a  point  in  its  plane  lies 
within  or  icithout  a  given  circumference,  according  as  its  dis- 
tance from  the  center  is  less  or  greater  than  the  radius. 


ANGLES. 

24.  A  plane  angle  is  the  opening  or  difference  of  direction 
between  two  lines  that  meet  or  might  meet.  The  point  of 
meeting  is  called  the  vertex;  and  the 
lines,  the  sides  of  the  angle. 

Thus  the  lines  AB,  AC,  meeting  in 
the  vertex  A,  are  said  to  form  or  include 
the  angle  called  BAC,  CAB,  or  sim- 
ply the  angle  A  when  no  other  angle 
has  the  same  vertex.  Even  when  sev- 
eral angles  have  a  common  vertex,  it 
is  sometimes  convenient  to  designate 
each  by  a  number  or  letter  placed 
within,  near  the  vertex.  Thus  the 
angles  MAR,  RAP,  PAN,  may  be  more  briefly  designated  as 
the  angles  1,  2,  and  A,  respectively. 


16 


PLANE   GEOMETRY. 


25.  Angles  are  equal  if  their  sides  can  be  made  to  coincide. 
Thus  the  angles  BAC  and  EDF 

are  equal  if,  DE  being  made  to  lie 
on  AB,  DF  can  also  be  made  to  lie 
on  AC,  which  is  possible  only 
when  the  opening  between  DE 
and  DF  is  the  same  as  the  opening  between  AB  and  AO. 

26.  Adjacent  angles  are  such  as  have  a  common  vertex  and 
a  common  side  that  separates  them.  Thus  the  angles  1  and 
2  in  Art.  24,  or  2  and  A,  are  adjacent  angles. 

Angles  that  have  a  common  side,  as  D  and 
C,  but  separate  vertices,  are  sometimes  called 
adjacent.  It  is  better,  however,  to  avoid  pos- 
sible confusion  by  calling  them  including  angles, 
as  including  the  common  side  between  their  vertices. 

27.  Vertical  angles  are  the  opposite  angles 
formed  by  the  intersection  of  two  straight 
lines.  Thus  the  angles  1  and  2  are  vertical 
angles,  as  are  also  the  angles  3  and  4. 

28.  The  sum  of  two  angles  is  equal  to  the  angle  formed 
by  placing  them  so  as  to  be  adjacent. 
The   difference   of  two   angles   is   the 
angle  that  added  to  the  smaller  angle 
makes  an  angle  equal  to  the  greater. 

Thus  BAD  is  the  sum  of  BAC  and 
GAD ;  BAC  is  the  difference  of  BAD  and  CAD. 

29.  A  straight  angle  is  one  whose  sides  lie  in  opposite 
directions  from  the  vertex,  so  as  to  P 
be  in  a  straight  line.                                                        ....•'•"■' 

Thus  5^c  is  a  straight  angle  if ..•■•"" . 

its  sides  AB,  AC  are  in  a  straight    ^  ^  ^ 

line.  It  may  be  regarded  as  formed  by  two  lines  drawn  in 
opposite  directions  from  A;  or  by  opening  out  a  smaller 
angle,  BAD,  until  its  sides  lie  in  opposite  directions;  or  as  the 


INTRODUCTION.  17 

sum  of  two  adjacent  angles,  BAD,  DAC,  whose  exterior  sides 
lie  in  a  straight  line. 

30.  When  the  adjacent  angles  formed  by  one  straight  line 
meeting  another  are  equal,  each  angle  is 
called  a  right  angle. 

Thus,  if  the  angles  BAC,  CAD,  formed 
by  CA  meeting  BD  in  A,  are  equal,  each  of 
them  is  a  right  angle.  d 

31.  A  right  angle  is  equal  to  one  half  of  a  straight  angle. 
This  follows  from  the  preceding  definitions ;  for  the  two 

equal  angles,  CAB,  CAD,  make  up  the  straight  angle  BAD. 

32.  An  acute  angle  is  less  than  a  right  angle ;  an  obtuse 
angle  is  greater  than  a  right  angle  and  less 
than  a  straight  angle.     Thus  BAC  is  an 
acute,  and  cad  an  obtuse,  angle. 

Acute  angles  and  obtuse  angles  are  called 
oblique  angles;  and  lines  are  said  to  be  j9er-  dab 
pendicular  or  oblique  to  each  other  according  as  they  meet 
at  a  right  or  an  oblique  angle. 

33.  If  the  sum  of  two  angles  is  equal  to  a  right  angle, 
each  is  called  the  complement  of  the  other,  and  the  angles 
are  said  to  be  complementary.  If  the  sum  of  two  angles  is 
equal  to  a  straight  angle,  each  is  called  the  supplement  of 
the  other,  and  the  angles  are  said  to  be  supplementary. 

QUESTIONS. 

1.  In  a  line  AB,  take  any  point  C ; 

then  AB  is  the  sum  of  what  two  lines  ?     '■ 1 ~ 

AC  IS  the  difference  of  what  two  ? 

2.  If  you  fold  a  piece  of  note  paper  so  as  to  form  an  edge,  what  sort 
of  a  line  is  formed  ? 

3.  If  you  fold  the  paper  again,  so  as  to  double  that  edge  upon 
itself,  what  angle  will  the  second  edge  thus  formed  make  with  the  lirst? 

Geom.— 2 


18  PLANE  GEOMETRY. 

4.  If  you  suspend  a  weight  by  a  string,  in  what  sort  of  a  line  is 
the  string  stretched  ? 

5.  If  you  whirl  the  weight  round  at  the  end  of  the  string,  in  what 
sort  of  a  line  does  the  weight  move  ? 

6.  What  sort  of  a  surface  is  presented,  roughly  speaking,  by  the 
walls  of  a  room  ?  By  the  surface  of  a  floor  ?  By  the  surface  of  a 
slate  ?     Mention  other  like  surfaces. 

7.  How  would  you  apply  a  straightedge  or  ruler  so  as  to  ascertain 
whether  a  given  surface  is  a  plane  ?  What  property  of  planes  do  you 
apply  (Art.  9)  ? 

8.  Straight  lines  can  be  drawn  on  the  surface  of  a  stovepipe,  and 
yet  it  is  not  a  plane  :  why  not  ? 

9.  Can  a  straight  line  be  drawn  on  the  surface  of  an  eggshell? 
If  not,  what  kind  of  a  line  can  be  drawn  on  such  a  surface  ? 

10.  From  a  point  0  draw  lines  OA^  OB,  OC,  OD,  in  one  plane. 
Name  each  of  the  angles  thus  formed.  Which  are  adjacent  angles  ? 
Name  one  that  is  the  sum  of  two  ;  of  three. 

11.  Can  you  draw  two  angles  that  have  a  common  vertex  and  a 
common  side,  and  yet  are  not  adjacent  ? 

12.  What  sort  of  an  angle  is  less  than  its  supplement  ?  Is  equal  to 
its  supplement  ?  Is  greater  than  its  supplement  ? 

PROPOSITIONS. 

The  truths  of  geometry  are  presented  for  consideration 
under  the  form  of  general  statements  called  propositions. 

34.  A  theorem  is  a  proposition  stating  a  geometrical  truth. 

35.  A  problem  is  a  proposition  stating  a  proposed  con- 
struction. 

36.  A  corollary  is  a  theorem  that  follows  so  plainly  as  a 
consequence  from  a  preceding  proposition,  or  definition, 
that  its  formal  proof  is  either  omitted  or  is  merely  indicated. 

Thus  in  Arts.  15,  16,  17,  are  given  certain  important 
corollaries  from  the  definition  of  the  straight  line;  and  in 
Art.  23  we  have  a  very  obvious  corollary  from  the  definitions 
of  circumference  and  radius. 


INTRODUCTION.  19 

37.  An  axiom  is  a  theorem  assumed  as  self-evident. 

38.  A  postulate  is  a  problem  assumed  as  possible. 

39.  A  scholium  is  a  remark  upon  a  preceding  proposition. 

40.  The  axioms  and  postulates,  together  with  the  defini- 
tions, constitute  the  logical  basis  of  geometry.  Axioms 
express  certain  simple  truths  in  regard  to  magnitude  in 
general,  —  truths  so  confirmed  by  all  our  experience  that 
the  mind  cannot  conceive  their  opposites  as  true.  All  the 
axioms  except  the  last  two,  which  are  really  definitions  of 
the  terms  whole  and  part,  might  be  deduced  from  the  first. 
They  are  such  obvious  truths,  however,  that  it  is  deemed 
sufficient  to  state  them  for  convenience  of  reference. 

AXIOMS. 

1.  Magnitudes  equal  to  the  same  or  equal  magnitudes  are 
equal  to  each  other. 

2.  If  equals  are  added  to  equals,  the  sums  are  equal 

3.  If  equals  are  taken  from  equals,  the  remainders  are 
equal. 

4.  If  equals  are  added  to  unequals,  the  sums  are  unequal. 

5.  If  equals  are  taken  from  unequals,  or  unequals  from 
equals,  the  remainders  are  unequal. 

6.  The  doubles  of  equals  are  equal. 

7.  The  halves  of  equals  are  equal. 

8.  The  whole  is  greater  than  any  of  its  parts. 

9.  The  whole  is  equal  to  the  sum  of  all  its  parts. 

POSTULATES. 

Let  it  be  assumed  that,  in  a  given  plane, 

1.  A  straight  line  can  he  drawn  joining  aity  two  given 
points. 

2.  A  given  straight  line  can  be  produced  to  any  extent. 


20  PLANE   GEOMETRY. 

3.  On  the  greater  of  two  straight  lines  a  part  can  he  laid 
off  equal  to  the  less. 

4.  A  circumference  can  he  described  from  any  center,  with 
any  radius. 

5.  A  figure  can  he  moved  unaltered  to  a  new  position. 

It  is  assumed  in  these  postulates  that  we  have  at  our 
disposal,  (1)  a  plane  extending  indefinitely  in  all  direc- 
tions, (2)  some  means  of  causing  a  marking  point  to  move 
in  a  straight  line  in  any  given  direction,  (3)  some  means 
of  causing  a  marking  point  to  move  in  that  plane  so  as 
always  to  remain  at  a  given  distance  from  a  given  point 
in  it.  The  plane  may  be  represented  by  a  blackboard  or  flat 
piece  of  paper ;  the  second  requirement  is  met  by  the  use 
of  a  straightedge  and  marking  point ;  the  third,  by  a  pair 
of  compasses,  or  other  device.  By  means  of  Post.  5,  con- 
taining the  principle  of  superposition,  we  are  enabled  to 
apply  the  criterion  of  Art.  14  in  order  to  prove  the  equality 
of  two  given  magnitudes.  Thus  two  straight  lines  would 
be  proved  equal  by  placing  them  so  as  to  coincide  end  with 
end;  two  angles,  by  causing  their  sides  to  coincide;  and 
so  on. 

METHOD    OF    PR00F.=* 

In  general,  the  statement  and  proof  of  a  proposition  con- 
sist of  several  distinct  parts ;  the  enunciation,  the  construc- 
tion, and  the  demonstration. 

1.  The  General  Enunciation  or  statement  consists 
of  an  hypothesis  (or  supposition)  and  a  conclusion.  Thus 
in  Prop.  I.  we  have,  though  in  different  words :  — 

Hypothesis.  If  a  line  is  perpendicular  to  a  second  line  at 
a  certain  point, 

*  To  be  read  in  connection  with  Prop.  I. 


INTRODUCTION.  '      21 

Conclusion.  No  other  line  in  the  same  plane  can  be  per- 
pendicular to  the  second  line  at  that  point. 

2.  The  Particular  Enunciation,  again,  refers  us  to 
a  particular  figure  or  figures  fulfilling  the  given  conditions. 
The  hypothesis  and  conclusion  of  the  particular  enunciation 
will  be  distinguished  by  the  headings  Given,  and  To  Prove, 
respectively. 

3.  In  the  Construction  we  apply  the  postulates,  or 
problems  that  have  been  proved  possible,  to  make  such 
changes  in  the  form  or  position  of  the  given  figures  as  may 
be  needful  for  the  demonstration. 

4.  In  the  Demonstration  we  deduce,  by  a  train  of 
reasoning,  the  proposition  to  be  proved,  from  other  propo- 
sitions already  proved  or  granted.  Thus,  in  Prop.  I.,  we 
show,  by  means  of  Ax.  8  and  Ax.  1  and  the  definition  of 
right  angles,  that  the  angles  formed  by  the  line  AE  with 
BD  must  be  unequal,  and  therefore  cannot  be  right  angles. 

In  problems,  the  construction,  not  always  a  necessity  in 
the  proof  of  a  theorem,  is  the  essential  part.  Instead  of 
the  heading  To  Prove,  however,  we  put  the  heading  Required, 
as  indicating  what  is  required  to  be  done. 

QUESTIONS. 

1.  What  kind  of  a  surface  and  what  instruments  are  assumed  as 
necessary  in  the  constructions  of  plane  geometry  ? 

2.  How  do  you  draw  a  straight  line  longer  than  your  straightedge  ? 
What  property  of  straight  lines  do  you  apply  ? 

3.  With  what  instrument  do    a B 

you  lay  off  on  a  line  AB  a  shorter 

line  C2>? 

4.  Draw  a  line  equal  to  the  given  line  AB ;  produce  it  to  E,  so 
that  BE  shall  be  equal  to  the  line  CD :  AE  is  the  sum,  and  BE  the 
difference,  of  what  lines  ? 


22  PLANE   GEOMETRY, 

5.  The  sum  of  a  right  angle  and  an  acute  angle  is  necessarily  what 
sort  of  an  angle  ? 

6.  The  difference  of  an  obtuse  angle  and  a  right  angle  is  neces- 
sarily what  sort  of  an  angle  ? 

7.  In  the  annexed  diagram,  show  that    ^  ^ 
there  are  three  different  angles  having  the 
same  vertex  A.     Name  them. 

8.  If  BA  were  produced  through  A  to  E,  how  many  different 
angles  would  have  the  same  vertex  A?    Name  them. 

Kead  the  following  expressions : 

9.  ZBAC+ZGAD  =  ABAD.  11.  ABAD>/.CAD. 
10.  A  BAD  -  /.B  AC  =  ACAD.  12.  A  BAG  <  A  BAD. 
Express  in  symbols  each  of  the  following  statements : 

13.  The  sum  of  the  lines  AB 

and    BC   is  equal   to   the  line    ^  b C 

AC.  »- '  ^ 

14.  The  difference  of  the  lines  AC  and  BC  is  equal  to  the  line  AB. 

15.  The  line  AC  i^  greater  than  the  line  BC. 

16.  The  line  AB  is  less  than  the  line  A  C. 

17.  The  sum  of  the  angles  BAD  and  BAG  is  a 
right  angle. 

18.  The  difference  of  the  angles  BAD  and  BAG  is  equal  to  the 
angle  GAD. 


Book  I. 


LINES  AND  RECTILINEAR  FIGURES, 


ANGLES. 

Proposition  I.     Theorem. 

41.  At  a  given  point  in  a  straight  line  there  can 
he  in  the  same  plane  but  one  perpendicular  to  that 
line. 

c      ,E 


A  D 

F 

Given:  CF  perpendicular  to  5D  at  ^; 

To  Prove :  No  other  line  drawn  through  A  can  be  X  to  BD. 

For  any  other  line  drawn  from  A  must  lie  between  FC  and 
AB  OY  between  FC  and  AD.  Let  it  lie  between  FC  and  AD, 
as  AE. 

Because  ZcAB  =  ZcADj  (Hyp.) 

but  Z  EAB  >  Z  CAB,   and   Zead<Z  cad,      (Ax.  8) 
Z  EAD  is  not  equal  to  Z  EAB,  (Ax.  1) 

.-.  EA  is  not  perpendicular  to  BD.     q.e.d.    (30) 
23 


24  PLANE   GEOMETRY.  — BOOK  I. 

42.    Cor.  1.  All  right  angles  are  equal. 

For  otherwise  two  could  be  so  placed,  as  in 
the  figure,  that  there  would  be  two  perpendicu- 
lars to  the  same  line  at  the  same  point. 


43.'    CoR.  2.  All  straight  angles  are  equal.  (Ax.  6) 

For  each  is  the  sum  of  two  right  angles.  (31) 

44.  CoR.  3.  The  complements  and  supplements  of  equal 
angles  are  equal,  and  conversely.*  (Ax.  3) 

45.  Definition.  If  two  theorems  are  so  related  that  the 
hypothesis  and  conclusion  of  the  one  are  respectively  the 
conchision  and  hypothesis  of  the  other,  the  one  theorem  is 
said  to  be  the  converse  of  the  other.  Thus,  as  will  presently 
be  seen,  Prop.  II.  and  Prop.  III.  are  converse  propositions. 
The  converse  of  a  proposition  is  by  no  means  necessarily 
true,  and,  if  true,  requires  to  be  established  by  demonstration. 

NoTK.  —  In  Prop.  I.,  as  in  all  the  propositions  of  Plane  Geometry, 
it  is  understood  that  all  the  points  of  each  figure  lie  in  one  and 
the  same  plane.  It  will  afterwards  be  seen  that  there  may  be  any 
number  of  lines,  each  perpendicular  to  a  given  line  at  the  same  point, 
but  not  all  in  the  same  plane  with  each  other  and  the  line. 

Exercise  1.  If  an  angle  is  double  its  complement,  what  fraction 
is  it  of  a  right  angle  ?    Of  a  straight  angle  ? 

2.  If  an  angle  is  three  times  its  supplement,  what  fraction  is  it  of 
a  straight  angle  ?    Of  a  right  angle  ? 

3.  The  lines  that  bisect  adjacent  complementary  angles  form  half  a 
right  angle. 

4.  In  the  diagram  for  Prop.  I.,  name  the  complements  of  the 
angles  CAE  and  EAD,  respectively. 

5.  In  the  same  diagram,  name  the  supplements  of  the  angles  EAD, 
EAB,  and  CAB.,  respectively.  The  sum  of  what  two  of  those  angles 
is  equal  to  the  supplement  of  angle  CAE  9 

6.  In  the  same  diagram,  suppose  EA  produced  to  6.  Show  that 
BAb  and  CAE  are  complementary  angles. 

*  Conversely  means  the  converse  proposition  is  also  true.    See  Art.  45. 


ANGLES. 


25 


Proposition  II.     Theorem. 

46.  If  two  adjacent  angles  have  their  exterior  sides 
in  the  same  straight  line,  the^e  angles  are  supple- 
mentary. 


Given:  Two  adjacent  angles  BAC,  cad,  with  their  exterior 
sides  AB,  AD,  in  a  straight  line  ; 

To  Prove:  BAC  and  CAD  are  supplementary  angles. 

Since  AB  is  in  a  straight  line  with  AD,        (Hyp.) 

BAD  is  a  straight  Z.  (29) 

But  Z  BAC  +  Z  CAD  =  Z  BAD,  (28) 

.*.  BAC  and  CAD  are  supplementary  A.    q.e.d.    (33) 

47.    Cor.  1.    The  sum  of  all  the  angles   formed  by  any 
number  of  lines  meeting  AB,from  the  same 
side,  in  a  point  O,  is  equal  to  a  straight 


angle  or  two  right  angles. 


(Ax.  8) 


48.  Cor.  2.  The  sum  of  all  the  angles 
formed  by  any  number  of  lines  meeting 
m  a  point  0  is  equal  to  two  straight  an- 
gles or  four  right  angles. 

For  if  any  of  the  lines  be  produced  through  0,  the  sum 
of  the  angles  on  each  side  of  that  line  will  be  a  st.  Z,  (47). 

Exercise  7.  In  the  diagram  for  Prop.  II.,  if  angle  BAC  is  equal 
to  twice  angle  CA  D,  what  fraction  is  each  of  a  right  angle  ? 

8.  If  four  lines,  OA,  OB,  OC,  OD,  meet  in  a  point  O,  so  that 
angle  ^  OB  wangle  COD,  and  angle  ^02>  =  angle  BOC,  then  OA 
and  OC  are  in  the  same  straight  line,  as  are  also  OB  and  OD;  and 
the  pairs  of  angles  AOB,  AOD,  and  BOC,  DOC,  are  supplementary. 


26  PLANE   GEOMETRY.  — BOOK  I. 

Propositiox  III.     Theorem. 

49.  If  two  adjacent  angles  are  supplementary,  their 
exterior  sides  are  in  a  straight  line. 


Given:  Two  adjacent  A,  BAC,  CAD,  that  are  supplementary; 
To  Prove :  Their  exterior  sides,  AB,  AD,  are  in  a  straight  line. 

Since  the  A  BAC,  CAD,  are  supplementary,    (Hyp.) 
Z  BAC  +  A  CAD  =  a  straight  A,  (33) 

.-.  their  sum,  A  BAD,  is  a  straight  A, 
.*.  AB  and  AD,  the  sides  of  A  BAD,  are  in  a  st.  line.    (29) 

Q.E.D. 


Proposition  IV.     Theorem. 

50.   If  two  straight    lines   intersect,    the    vertical 
angles  are  equal. 


-D 

Given:  Two  straight  lines,  AB,  CD,  intersecting  in  o; 
To  Prove:  Angle  ^oc  is  equal  to  angle  BOD,  and  angle  AOD 
is  equal  to  angle  BOC. 

Since  A  aod  is  supp.  to  A  AOC,  and  also  to  A  BOD,     (46) 
A  AOC  =  A  BOD.  (44) 

Similarly,    A  AOD  =  A  BOC, 
each  being  the  supplement  of  A  BOD.  q.e.d. 


ANGLES.  27 

Proposition  V.     Theorem. 

51.  From  a  point  without  a  line,  there  can  he  hut 
one  perpendicular  to  that  line. 


Given:  The  point  A,  the  line  BC,  AB  A.  to  BC,  and  AC 
any  other  line  from  A%o  BC; 

To  Prove :  ^C  is  not  perpendicular  to  BC. 

Turn  the  figure  ABC  about  BC  till  A  takes  the  position  A^ 
in  the  original  plane.  Mark  the  point  A\  restore  ABC  to 
its  first  position,  and  join  A^B,  a'c. 

Then  since  AB  and  AC  can  be  made  to  coincide  with  a'b 

and  a'c,  while  BC  retains  its  position,      (Const.) 

Z  ABC  =  Z  a'bc,   and  Z  ACB  =  Z  a'cb.       (25) 

But  Z  ABC  is  a  right  Z,  (Hyp.) 

.-.  Z  ^'5C  is  a  right  Z,  (30) 

.'.  aba'  is  a  straight  line,  (49) 

.*.  ^C^'  is  not  a  straight  line,  (6) 

.-.  Z  AC  A'  is  not  a  straight  Z,  (29) 

.*.  Z  ^C-B,  the  half  of  Z  ^c^',  is  not  a  right  Z.     q.e.d. 

Exercise  9.  If  in  the  diagram  for  Prop.  III.,  angle  CAD  is  |  of  a 
right  angle,  what  angle  must  BA  make  with  ^C  so  that  BA  shall  be 
in  the  same  straight  line  with  AD  ? 

10.  If,  in  a  line  MN,  a  point  P  be  taken,  and  two  lines  PQ,  Plt^ 
be  drawn  so  that  angle  QPM  =  angle  BPN^  then  QB  is  a  straight  line. 

11.  In  the  diagram  for  Prop.  V.,  the  angles  A  and  A'  are  the  sup- 
plements of  what  angles  ?     Show  that  these  angles  are  equal. 

12.  In  this  diagram,  show  that  the  exterior  angles  at  C  are  equal. 


28 


PLANE  GEOMETRY.  — BOOK  I. 


TRIANGLES. 

52.  A  triangle  is  a  portion  of  a  plane  bounded  by  three 
straight  lines. 

53.  The  lines  that  bound  a  triangle  are  called  its  sides; 
the  angles  formed  by  the  sides,  its  angles;  and  the  vertices 
of  the  angles,  the  vertices  or  angular  points  of  the  triangle. 

If  all  the  sides  of  a  triangle  be  produced  both  ways  (see 
triangle  C  below),  nine  new  angles  will  be  formed  in  addi- 
tion to  those  properly  called  the  angles  of  the  triangle, 
or  by  way  of  distinction,  its  interior  angles.  Of  the  nine 
outer  angles,  the  six  angles  that  are  supplementary  to  the 
interior  angles,  are  called  exterior  angles.  Interior  angles 
are  always  meant  whenever  we  refer  to  the  angles  of  a 
triangle  without  any  distinguishing  epithet. 

54.  The  base  of  a  triangle  is  the  side  on  which  it  is  sup- 
posed to  stand ;  the  angle  opposite  the  base  is  sometimes 
referred  to  as  the  vertical  angle. 


55.  An  equilateral  triangle  has  three  equal  sides ;  as  A. 

56.  An  isosceles  triangle  has  two  equal  sides ;  as  ^. 

57.  A  scalene  triangle  has  no  two  sides  equal ;  as  C. 


TRIANGLES.  29 

58.  A  right  triangle  has  a  right  angle ;  as  D. 

59.  An  obtuse  triangle  has  an  obtuse  angle ;  as  E. 

60.  An  acute  triangle  has  all  its  angles  acute ;  as  F. 

While  lines  and  angles  can  be  equal  only  in  one  way,  it 
is  different  in  regard  to  triangles  and  other  inclosed  figures. 
For  as  will  afterwards  be  seen,  two  triangles  that  are  equal 
as  regards  surface  may  differ  greatly  as  to  their  sides  and 
angles.  Hence  it  is  necessary  to  define  the  term  equal  in 
regard  to  figures  in  general. 

61.  Equal  figures  are  such  as  can  be  made  to  coincide 
exactly ;  that  is,  are  equal  in  every  respect. 

62.  Theorem.  Two  triangles  are  equal  if  their  angular 
points  can  he  made  to  coincide. 

For  if  the  angular  points  coincide,  the  sides  terminated 
by  those  points  must  coincide  (14) ;  hence,  also,  the  angles 
formed  by  the  sides,  and  the  surfaces  bounded  by  them, 
must  coincide.  It  is  evident  that  the  theorem  may  be 
extended  so  as  to  apply  to  figures  bounded  by  any  number 
of  straight  lines,  the  reasoning  being  exactly  the  same. 

Exercise  13.  Show,  by  Prop.  V.,  that  no  triangle  can  have  two 
of  its  angles  right  angles. 

14.  If  triangle  A  has  all  its  angles  equal,  show  that  its  six  exterior 
angles  are  all  e(iual. 

15.  If  a  triangle  has  two  equal  angles,  those  angles  must  be  acute. 
What  about  the  third  angle  ? 

16.  If  a  triangle  has  no  two  angles  equal,  it  has  three  different  pairs 
of  equal  exterior  angles. 

17.  If,  in  triangle  Z),  the  sides  containing  the  right  angle  be  pro- 
duced through  its  vertex,  the  sum  of  the  exterior  angles  thus  formed 
is  equal  to  a  straight  angle. 

18.  If,  in  triangle  E,  the  sides  containing  the  obtuse  angle  be 
produced  through  its  vertex,  the  sum  of  the  exterior  angles  thus 
formed  is  less  than  a  straight  angle. 


30  PLANE   GEOMETRY.  —  BOOK  I. 

Proposition  YI.     Theorem. 

63.  Two  triangles  are  equal  if  a  side  and  the 
including  angles  of  the  one  are  respectively  equal  to 
a  side  and  the  including  angles  of  the  other. 


Given:  In  triangles  ABC,  a'b'c',  bc  equal  to  B'c',  angle  B 
equal  to  angle  B',  and  angle  C  equal  to  angle  c' ; 

To  Prove:  Triangle  ABC  is  equal  to  triangle  a'b'c'. 

liAABC  be  placed  upon  A  a'b'c',       (Post.  5) 
so  thsit  BC=^  B'c', 
then  since  Z  B  =  Z  b',  (Hyp.) 

BA  will  take  the  direction  of  b'A',  (25) 

and  A  will  lie  on  b'a'  ov  b'a'  produced. 

Also  since  Zc  =  Zc',  (Hyp.) 

CA  will  take  the  direction  of  c'a',  (25) 

and  A  will  lie  on  c'A'  or  c'a'  produced. 
Then  since  A  lies  on  both  b'a'  and  C'A\ 

A  must  coincide  with  A', 
the  only  point  common  to  b'a'  and  C'a',  (6) 

.-.  Aabc  =  Aa'b'c'.  q.e.d.     (62) 

64.  Scholium.  If  the  including  angles  are  all  equal,  i.e., 
if  Z  C  =  Z  B  =  Z  c'  =  Z  b',  as  in  the  right  hand  pair  of 
triangles,  it  is  manifest  that  A  ABC  can  be  made  to  coincide 
with  A  A'b'c'  in  the  reverse  position  also,  so  that  AB  will 
coincide  with  a'c',  and  AC  with  a'b'. 


OF  THE 

UNIVERSITY 

^^  ^  TRIANGLES.  81 


Proposition  VII.     Theorem. 

65.  If  two  angles  of  a  triangle  are  equal,  the  sides 
opposite  those  angles  are  also  equal. 


Given:  In  triangle  ABC,  angle  C  equal  to  angle  B ; 
To  Prove:  ^i?  is  equal  to  ^C. 

Turn  A  ABC  about  its  base  BC  till  A  takes  the  position 
A'  (Post.  5).  Mark  the  point  a\  restore  ABC  to  its  first 
position,  and  join  A^B,  A'C  (Post.  1),  so  as   to  form    the 

Aa'bc, 

Since  their  angular  points  can  be  made  to  coincide,  (Const.) 

AABC  =  AA'BC,  (62) 

and  A'B  =  AB,  A'C  =  AC,Z  a'bc  =  Zb,Z  A'CB  =  Zc.  (14) 

Since  these  four  angles  are  all  equal,     (Hyp.  and  Const.) 

A  ABC  can  be  made  to  coincide  with  Aa'bc 

in  the  reverse  of  the  first  position,  (64) 

so  that  AB  ^  A'C  and  AC  =^  a'b. 

Then  since  AB  =  A'c,  (14) 

and^C  =  ^'C,  (Above) 

ABz=AC.  Q.E.D.      (Ax.  1) 

Exercise  19.  In  the  diagram  for  Prop.  VIL,  if  AA'  be  drawn,  cut- 
ting BC  in  O,  show  that  when  triangle  ABC  is,  folded  over  on  triangle 
A'BC,  AO  must  coincide  with  A'O. 


32 


PLANE   GEOMETRY.  — BOOK  I. 


Proposition  VIII.     Theorem. 

66.  Two  triangles  are  equal  if  two  sides  and  the 
included  angle  of  the  one  are  respectively  equal  to 
two  sides  and  the  included  angle  of  the  other. 


Given:  In  triangles  ABC,  a'b'c',  AB  equal  io  A^B^,  AC  equal 
to  A'c\  and  angle  A  equal  to  angle  A' ; 

To  Prove:  Triangle  ABC  is  equal  to  triangle  a'b'c'. 

HAabc  be  placed  upon  A  a'b'c',       (Post.  5) 
so  that  Z  ^  r^  Z^', 
then  since  AB  =  A'b',  (Hyp.) 

B=^B'. 

Also  since  yl (7  =  ^'C',  (Hyp.) 

c  ^  c', 

.'.  A  ABC  =  A  a'b'c'.  q.e.d.     (62) 

67.  Scholium.  If  the  including  sides  are  all  equal,  i.e.,  if 
AB  —  AC  =  A'b'  =  a'c',  as  in  the  right  hand  pair  of  trian- 
gles, it  is  manifest  that  A  ABC  can  be  made  to  coincide  with 
A  a'b'c'  in  the  reverse  position  also,  so  that  B  will  coincide 
with  C',  and  C  with  B'. 

Exercise  20.  Show  (Exercise  19)  that  ^0  is  perpendicular  to  EC, 
bisects  angle  A,  and  bisects  base  BC. 

21.  Prove,  by  Prop.  VI.,  that  all  the  triangles  having  their  vertices 
at  0  are  equal. 

22.  Prove  the  same  equalities  by  means  of  Prop.  VIII. 

23.  In  the  triangles  ABC,  A'B'C,  above,  when  AB  is  made  to 
coincide  with  A'B',  where  would  AC  fall  (1)  if  angle  A  were  greater 
than  angle  A' ;  (2)  if  less  than  angle  A'f 


TRIANGLES.  33 

Proposition  IX.     Theorem. 

68.  If  two  sides  of  a  triangle  are  equal,  the  angles 
opposite  them  are  equal. 


Given :         In  triangle  ABC,  AC  equal  to  AB ; 
To  Prove :        Angle  B  is  equal  to  angle  C. 

Turn  A  ABC  about  its  base  BC  till  A  takes  the  position 
A'  (Post.  5).  Mark  the  point  A',  restore  ABC  to  its  first 
position,  and  draw  A^B,  A'c  (Post.  1). 

Since  their  angular  points  can  be  made  to  coincide,  (Const.) 

AABC  =  AA'BC,  (62) 

and  A'b  =  AB,  A'c  =  AC,  Z  a'bc  =  Zb,Z  A'CB  =  Z  C.  (14) 

Since  all  these  sides  are  equal,      (Hyp.  and  Const.) 

A  ABC  can  be  made  to  coinciile  with  A  A'BC 

in  the  reverse  of  the  first  position,  (67) 

so  that  Zb  =^  Z  A'CB,  and  Z  c  =^  Z  A'bc. 

Then  since  Zb  =  Z  A'CB,  (14) 

and  Zc  =  Z  A'CB,  (Above) 

Zb=Zc.  q.e.d.     (Ax.  1) 

Exercise  24.   If  two  sides  of  a  triangle  are  unequal,  the  angles 
opposite  to  them  are  also  unequal,  and  conversely. 
25.   A  scalene  triangle  cannot  have  two  equal  angles. 
Geom. — 3 


34  PLANE    GEOMETRY.  — BOOK  I. 

Proposition  X.     Theorem. 

69.  Two  triangles  are  equal  if  the  three  sides  of 
the  one  are  respectively  equal  to  the  three  sides  of 
the  other. 


A 

Given:  In  triangles  ABC,  A'b'c',  AB  equal  to  a'b\  AC  equal 
to  A'&,  and  BC  equal  to  B^& ; 

To  Prove:  Triangle  ABC  i&  equal  to  triangle  a'b'c'. 

Apply  A^^C  to  A  A'b'c', 

so  that  BC  =^  b'c',  and 

let  A  fall  on  the  further  side  of  B'c'  from  A'. 

Join  AA',  and  first  suppose  AA'  cuts  B'c'. 

Since  AB'  =  a'b',  and  AC'  =  A'c',  (Hyp.) 

Z  1  =  Z  2,  and  Z  3  =  Z  4,  (68) 

.-.  Z1  +  Z3  =  Z2  +  Z4,  (Ax.  2) 

i.e.,  Z.BAC  =  Z.B'A 'c', 

.'.  A  ABC  =  A  a'b'c'.  q.e.d.     (66) 

If  A  a'  should  pass  through  an  extremity  of  B'c',  or  cut 
b'c'  produced,  the  proof  would  be  similar  to  that  above. 
But,  as  will  afterwards  be  seen,  the  triangles  can  always 
be  applied  to  each  other  so  that  A  A'  shall  cut  B'c'  between 
B'  and  C'. 

70.  Scholium.  In  equal  triangles,  sides  that  are  equal 
to  each  other  subtend  equal  angles,  and  equal  angles  are 
subtended  by  equal  sides. 


TRIANGLES. 


35 


For  when  equal  triangles  are  made  to  coincide,  coinciding 
sides  necessarily  subtend  coinciding  angles. 

This  important  proposition  is  a  scholium  rather  than  a 
corollary,  because  it  does  not  deduce  a  new  truth,  but 
merely  calls  attention  to  a  truth,  or  set  of  truths,  already 
virtually  proved.     See  definition,  p.  19. 

71.  Definition.  In  a  right  triangle,  the  side  subtending 
the  right  angle  is  called  the  hypotenuse ;  and  the  other  sides 
are  called  arms. 


Proposition  XI.     Theorem. 

72.  Two  right  triangles  are  equal  if  the  hypotenuse 
and  an  arm  of  the  one  are  respectively  equal  to  the 
hypotenuse  and  an  arm  of  the  other. 


Given:  In  right  triangles  ABC,  A'b'c',  hypotenuse  AC  equal 
to  hypotenuse  A'c',  and  AB  equal  to  a'b' ; 

To  Prove :  Right  triangle  ABC  is  equal  to  right  triangle  A 'B 'c'. 


)ply  A  ABC  to  A  A'b'c',  so  that  AC  ^  A'C', 

and  B  falls  remote  from  b'.     Join  BB'. 

Since  AB  =  a'b', 

(Hyp.) 

Za'b'b==Za'bb', 

(68) 

.-.  Z  c'b'b  =  Z  c'bb'. 

(44) 

(being  complements  of  equal  A,) 

.:   BC'  OT  BC  =  B'C', 

(65) 

.'.  Aabc  =  aa'b'c'.            Q.E.D 

•     (69) 

36  PLANE   GEOMETRY.  — BOOK  I. 

Proposition  XII.     Theorem. 

73.  Two  right  triangles  are  equal  if  the  hypote- 
nuse and  an  acute  angle  of  the  one  are  respectively 
equal  to  the  hypotenuse  and  an  acute  angle  of  the 
other. 


Given:  In  right  triangles  ABC,  A^b'c',  hypotenuse  BC  equal 
to  hypotenuse  B'c',  and  angle  B  equal  to  angle  B' ; 

To  Prove :  Eight  triangle  ABC  is  equal  to  right  triangle  A'b'c'. 

Place  A  ABC  upon  A  A'b'c',  so  that  BC  =^  b'c'. 

Then  since  Z  B  =  Z  B',  (Hyp.) 

BA  will  take  the  direction  oi  b'a' ,  (25) 

and  A  will  fall  upon  A' ; 

since  there  can  be  but  one  ±  from  c'  to  b'A' ;     (51) 

.-.  AABC  =  AA'B'C'.  Q.E.D.     (62) 


The  student  is  now  in  possession  of  all  the  theorems  con- 
cerning equal  triangles  that  are  of  practical  importance, 
viz.  — 

Two  triangles  are  equal  if  they  have,  respectively  : 

1.  One  side  and  the  including  angles  equal;         Prop.  VI. 

2.  Two  sides  arid  the  included  avgle  equal;      Prop.  VIII. 

3.  Three  sides  wMually  equal.  Prop.  X. 

4.  If  right,  the  hypotenuse  and  an  arm,  or  the  hypotenuse 
and  an  acute  angle,  mutually  equal.  Props.  XI.,  XII. 

It  may  be  easily  proved  that  triangles  are  equal  that  have 
a  side  and  any  two  angles  equal ;  the  case  is,  however,  of  no 
importance,  nor  is  the  more  diiRcult  one  concerning  triangles 
having  two  sides  and  a  not  included  angle  mutually  equal. 


TRIANGLES.  37 

Proposition  XIII.     Theorem. 

74.  The  line  that  joins  the  vertices  of  two  isosceles 
triangles  on  the  same  base,  bisects,  1°,  the  vertical 
angles,  2°,  the  common  base  at  right  angles. 


Given:  A  A'  joining  the  vertices  of  isosceles  triangles  ABC, 
A'bc,  and  cutting  BC  in  B ; 

To  Prove :  A  A'  bisects  angles  A  and  A',  and  is  perpendicular  to 
BC  Sit  its  mid  point. 

1°.  As  AB  =  AC,  A'b  =  A'C,  and  AA'  is  common,=^  (Hyp.) 

Abaa'  =  Acaa',  (69) 

.-.  Z  BAD  =  Z  CAD,  and  Zba'd  =  Z  CA'd.     q.e.d.   (70) 

2°.  As  AB  =  AC,  AD  is  common,  and  Z  BAD=Z  CAD,  (1°) 

A  BAD  =  A  CAD,  (66) 

.'.  BD  =  CD,  and  the  A  Sit  D  are  equal,  (70) 

.-.  A  A'  is  ±  to  BC  Sit  its  mid  point,     q.e.d.   (30) 

75.  Cor.  1.  If  two  points  are  each  equidistant  from  the 
extremities  of  a  given  line,  the  line  joining  these  points  is 
perpendicular  to  the  given  line  at  its  midpoint. 

The  points,  it  will  be  observed,  may  be  on  different  sides 
of  the  given  line,  or  both  on  the  same  side. 

*  That  is,  is  common  to  (or  belongs  to)  both  triangles. 


38  PLANE   GEOMETRY.— BOOK  1. 

76.  Definition.  The  line  that  divides  an  angle  into  two 
equal  angles  is  called  its  bisector. 

77.  Cor.  2.  The  bisector  of  the  vertical  angle  of  an  isosceles 
triangle  bisects  the  base  at  right  angles;  and  conversely. 

Thus  far  the  constructions  required  as  aids  in  the  demon- 
stration of  theorems  have  called  for  nothing  beyond  the 
postulates.  We  shall  presently  have  occasion  for  other 
constructions,  such  as  bisection  of  lines,  angles,  etc.  The 
problem  about  to  be  given  is  of  some  importance  as  entering 
largely  into  subsequent  problems. 

Exercise  26.  In  the  diagram  for  Prop.  IX.,  show  that  angle  ABA'  = 
angle  AC  A'. 

27.  Prove,  by  means  of  Prop.  X.,  the  equality  proved  in  Exercise  22. 

28.  Draw  the  diagram  for  Prop.  X.  when  the  angle  B  is  obtuse, 
and  prove  the  proposition. 

29.  Prove  that  on  the  same  side  of  the  same  base  there  can  be  but 
one  isosceles  triangle  having  its  arms,  or  equal  sides,  eacli  equal  to  a 
given  line. 

30.  Prove  that  on  the  same  side  of  the  same  base  there  can  be  but 
one  isosceles  triangle  having  a  given  vertical  angle. 

31.  In  the  diagram  for  Prop.  XIII.,  prove  in  regard  to  both  figures 
that  the  angle  ABA'  is  equal  to  the  angle  AC  A'. 

32.  If  the  base  of  an  isosceles  triangle  ^I?C  be  produced  both  ways 
to  D  and  £',  so  that  BD  =  CE,  and  A  be  joined  with  D  and  E,  the 
triangle  ADE  will  be  isosceles. 

33.  If  the  exterior  angles  formed  by  producing  both  ways  a  side  of 
a  triangle  are  equal,  the  other  sides  are  equal. 

34.  Right  triangles  are  equal  if  their  arms  are  respectively  equal. 

35.  Right  triangles  are  equal  if  they  have  equal  acute  angles  at  the 
extremities  of  equal  arms. 

36.  In  the  diagram  for  Prop.  XIV.,  if  BA  be  produced  to  meet  the 
circumference  at  Z>,  show  that  DE  is  equal  to  the  sum  of  the  three 
sides  of  triangle  CAB. 

37.  In  the  same  diagram,  BA  being  produced  as  above,  and  C 
joined  with  D  and  E,  show  that  CAD,  CBE,  are  equal  isosceles 
triangles. 

38.  In  the  same  diagram,  if  the  circumferences  intersect  a  second 
time  at  F,  and  CF  cut  AB  in  G,  show  that  GE=3AG. 


TRIANGLES. 


39 


Proposition  XIV.     Problem. 

78.  To  find  a  point  equidistant  froin  the  extremi- 
ties of  a  given  line. 


A'L 


Given :  A  straight  line  AB  ; 

Required :     To  find  a  point  equidistant  from  A  and  Bx 


From  A  as  center,  with  radius  AB,  describe 

the  circumference  BCD. 
From  B  as  center,  with  radius  BA,  describe 
the  circumference  ACE. 

Produce  ^^  to  meet  ACE  in  E. 

Since  AE  >  AB, 

E  lies  without  the  circumf.  BCD. 

But  A  lies  within  the  circumf.  BCD, 

.'.  circumf.  ACE  must  intersect  circumf.  BCD. 

Let  them  intersect  in  c.     c  is  the  point  required. 

For  since  AC  =  AB^ 

and  BC  —  BA  or  AB, 

.'.   AC  =  BC,  (Ax.  1) 

'.  a  point  C  has  been  found  equidistant  from  A  and  B.    q.e.f. 


J 


Post.  (4) 

(Post.  2) 

(Const.) 

(23) 

(21) 


(23) 


79.  Scholium.  Although,  for  convenience  of  demonstra- 
tion, the  radii  have  been  taken  each  equal  to  the  given  line, 
it  is  manifest  that  other  points  equidistant  from  A  and  B 


40  PLANE   GEOMETRY.— BOOK  I. 

can  be  obtained  by  taking  any  equal  radii  greater  than  one 
half  the  given  line.  It  is  also  obvious  that,  in  practice, 
instead  of  entire  circumferences,  only  such  portions  need 
be  described  as  will  give  the  points  of  intersection. 


Proposition  XV.     Problem. 
80.  To  bisect  a  given  straight  line. 


i^ 


Given :  A  straight  hne  AB  ; 

Required :  To  bisect  AB  ;  that  is,  to  find  its  mid  point. 

Find  two  points  C  and  D,  each  equidistant  from  A  and  B.  (78) 
Join  CD.     CD  will  intersect  ^S  at  its  mid  point. 
Since  C  and  D  are  two  points  equidistant  from  A  and  B, 
CD  is  _L  to  AB  at  its  mid  point,  say  E  ;        (75) 
i.e.,  AB  is  bisected  in  E.  q.e.f. 

Scholium.    This  construction  gives  not  only  the  mid  point 
of  AB,  but  also  the  perpendicular  to  AB  through  its  mid  point. 

Exercise   39.    Find  a  point  X  that  shall  be  twice   as  far  from 
each  of  two  given  points,  A  and  B,  as  A  is  from  B. 

40.  Find  a  second  point  Y  that  shall  be  three  fourths  the  distance 
from  each  of  those  same  points,  A  and  JB,  that  A  is  from  B. 

41.  Join  AB,  and  prove  that  the  perpendicular  at  the  mid  point  of 
^B  will  pass  through  X  and  Y. 


TRIANGLES.  41 

Proposition  XVI.     Problem. 
81.  To  bisect  a  given  angle. 


c 


Given :  An  angle  BAC; 

Required:  To  bisect  angle  ^^C. 

On  AB,  AC,  lay  off  any  equal  parts  AD,  AE.       (Post.  3) 

Join  DE  (Post.  1),  and  find  a  point  F  equidistant  from  D 

and  E.  (78) 

The  line  joining  AF  is  the  bisector  of  the  given  angle. 

Since  A  and  F  are  each  equidistant  from  D  and  E,     (Const.) 

AF  is  ±  to  DE  at  its  mid  point,  (75) 

.-.  Z  ^^(7  is  bisected  by  ^-F.       q.e.f.     (77) 

82.  Scholium.  By  repeating  the  operations  described  in 
Props.  XV.  and  XVI.,  a  given  line  or  angle  can  be  divided 
into  4,  8,  16,  •••,  2",  equal  parts. 

Exercise  42.  In  the  diagram  above,  show  that  angle  BDE  =  angle 
CED. 

43.  Divide  a  given  angle  ABC  into  four  equal  angles. 

44.  In  the  diagram  above,  if  the  point  F  were  taken  on  the  same 
side  of  DE  as  A,  what  three  positions  might  F  have  in  regard  to  ^  ^ 
In  which  of  these  positions  would  the  point  taken  fail  to  assist  in  the 
required  construction  ? 

45.  In  the  diagram  above,  join  F  with  D  and  E,  and  prove  the 
proposition  by  Prop.  X. 

46.  Show  that  in  an  equilateral  triangle  the  three  lines  joining  the 
vertices  with  the  mid  points  of  the  opposite  sides  are  (1)  equal  to  each 
other  ;  (2)  perpendicular  to  these  sides  ;  (3)  bisectors  of  the  angles. 


42  PLANE   GEOMETRY.— BOOK  L 

Proposition  XVII.     Theorem. 

83.  An  exterior  angle  of  any  triangle  is  greater 
than  either  of  the  remote  interior  angles. 


Given:  ACD,  an  exterior  angle  of  triangle  ABC; 

To  Prove :  Angle  ACD  is  greater  than  angle  A  or  angle  B. 

Of  the  two,  let  Z  A  he  not  less  than  Z  J?.* 
Bisect  J.C  in  ^  (80) ;  join  BE,  and  produce  BE  to  F,  so 
that  EF  =  BE  (Posts.  1  and  3).     Join  FC. 


Then  since  EA  =  EC,  and  EF  = 

zEB, 

(Const.) 

also  Z  AEB  =  Z  CEF, 

(50) 

Aeab  =  Aecf, 

(66) 

.'.  Zecf=  Za. 

(70) 

But  Zacd>  Zecf, 

(Ax.  8) 

.'.   Z  ACD  >  Z  A  ; 

and  since  Zb  is  not  greater  than  Z  A, 

(Hyp.) 

Z  ACD  >  Z  i?,  as  well  as  >  Z  ^. 

Q.E.D. 

84.  Cor.  1.  No  two  angles  of  a  triangle  can  be  supple- 
mentary. For  each  is  less  than  the  exterior  angle  which  is 
the  supplement  of  the  other. 

85.  Cor.  2.  If  a  triangle  has  an  obtuse  or  a  right  angle, 
each  of  the  others  is  acute. 

*  That  is,  angle  B  is,  at  most,  equal  to  angle  A. 


TRIANGLES.  43 

Proposition  XVIII.     Theorem. 

86.  A  greater  side  of  a  triangle  subtends  a  greater 
angle 


Given :  In  triangle  ABC,  BC  greater  than  AC ; 

To  Prove :        Angle  A  is  greater  than  angle  B. 

Produce  CA  to  n,  so  that  en  =  cb  ;  and  join  BD. 

Since  CB  =  CD,  (Const.) 

Zd  =  Zdbc.  (68) 

But  Za>Zd,  (83) 

(angle  A  being  an  exterior  Z  of  A  DAB,) 

.-.  Z  A  >  Zdbc, 

.-.  Za>Zb.  q.e.d.     (a.f.)* 

87.  Cor.  Conversely,  a  greater  angle  is  subtended  by  a 
greater  side. 

In  A  ABC  suppose  Z  A>  Z  B.  AC  cannot  be  equal  to  J5  c  ; 
for  then  Z  B  would  be  equal  to  Z  A  (65) :  nor  can  ^C  be 
greater  than  BC ;  for  then  Z  B  would  be  greater  than  Z  A 
(86).  Hence,  since  BC  can  neither  be  equal  to  nor  less 
than  AC,  it  must  be  greater  than  AC.     q.e.d. 

Exercise  47.  Draw  the  diagram  for  Prop.  XVII.,  when  the  exterior 
angle  ACD  is  acute. 

48.  Prove  that  a  perpendicular  from  a  vertex  to  the  opposite  side  of 
a  triangle  falls  (1)  within  the  triangle  if  the  angles  including  that  side 
are  both  acute  ;  (2)  without  the  triangle  if  one  of  those  angles  is  obtuse. 

*  If  m  >  n  and  n  >p,  then  a  fortiori  (i.e.,  with  greater  reason),  m  >p. 
This  is  called  the  argument  a  fortiori,  abbreviated  a.f. 


44  PLANE   GEOMETRY.  — BOOK  I. 

Proposition  XIX.     Theorem. 

88.   Any  side  of  a  triangle  is  less  than  the  sum  of 
the  other  two. 


driven:  Any  side  jBC  of  a  triangle  ABC; 

To  Prove :  BCis  less  than  AB  -{-  AC. 

Produce  BA  to  D,  so  that  AD  =  AC ;  and  join  DC. 

Since  AC  =  AD,  (Const.) 

/.acd  =  Zd.  (68) 

But  Zbcd>  Zacd,  (Ax.  8) 

.-.  Z.BCD  >Zd, 

.'.  BD,  or  BA  -i-  AC,>BC.       Q.E.D.         (86) 

89.  Cor.  Any  side  of  a  triangle  is  greater  than  the  dif- 
ference of  the  other  two. 

¥oT  since  BC-\- AC>AB,  (88) 

BC>AB  —AC.  (Ax.  5) 

Exercise  49.  Prove  Prop.  XVIII.  by  cutting  off  on  BC,  CD  equal 
to  CA,  joining  AD,  etc. 

60.  The  perpendicular  to  the  greatest  side  from  the  opposite  vertex 
falls  within  the  triangle. 

51.  Prove  Prop.  XIX.  by  supposing  BO  to  be  the  greatest  side,  and 
drawing  AD  perpendicular  to  BC. 

52.  In  a  triangle  ABC,  having  AB  less  than  AC,  prove  that  any 
point  in  the  line  joining  A  with  the  mid  point  ot  BC  is  nearer  to  B 
than  to  C. 


TRIANGLES. 


45 


Proposition  XX,     Theorem. 

90.  //  two  triangles  have  two  sides  of  the  one 
respectively  equal  to  two  sides  of  the  other,  hut  the 
included  angles  unequal,  the  greater  angle  is  sub- 
tended hy  a  greater  base. 


Given:   Two  triangles  ABC,  A'b'c',  having  AB  ec[ual  to  A'b', 
AC  equal  to  A'c',  but  angle  A  greater  than  A' ; 
To  Prove :    Base  BC  is  greater  than  base  b'c'. 

Place  A  a'b'c'  upon  A  ABC,  so  that  a'b'  =^  AB.     Then 
•.•  Za>Za'  (Hyp.),  ^'C'' will  fall  between  ^5  and^c,  as^c'. 
Draw  AD  bisecting  Z  CAC'  to  meet  BC  in  D,  and  join  DC'. 
Since  AC'  =  AC  (Hyp.),  and  AD  is  common, 

also  Zdac'  =  Zdac,  (Const. ) 

.-.  AADC'  =  AADC  (66),  3ind  DC' =  DC.  (70) 

But  BD-\- DC' >BC',  (88) 

and  BD  -^DC  =  BC,  (Const.) 

.*.  BC  >  BC'  or  b'c'.  q.e.d. 

91.  Cor.  Conversely,  if  triangles  ABC,  A'b'c',  have  AB, 
AC,  equal  to  a'b',  A'c',  respectively,  hut  the  base  BC  greater 
than  the  base  B'c',  then  the  angle  A  is  greater  than  the  angle  A'. 

For  Z  A  cannot  be  equal  to  Z.A',  since  then  BC  would  be 
equal  to  B'c'  (66);  nor  can  Z  A  he  less  than  Z  A',  since 
then  BC  would  be  less  than  B'c'  (90).  Hence  Z  A  must  be 
greater  than  Z  A'. 


46 


PLANE   GEOMETRY.  — BOOK  I. 


PERPENDICULARS. 

Proposition  XXI.     Problem. 

92.    To  draw  a  perpendicular  to  a  given  line  from 
a  given  point  without  it. 


F  .■ 


■■■-..  G 


D    .         H         .-E 

"■■ -'C 

Given :  A  point  P  without  a  straight  line  AB  of  indefinite  length ;  * 
Required:  To  draw  a  perpendicular  from  P  to  AB. 

Take  any  point  C  on  the  side  of  AB  remote  from  P. 

From  P  as  center,  with,  radius  PC,  describe  a  circumfer- 
ence, CFG.  (Post.  4) 

Since  AB  extends  indefinitely  between  P  and  C,  and  the 
circumference  may  be  described  in  either  direction  from  C, 
it  will  intersect  AB  in  two  points,  D,  E. 

Bisect  DE  in  H  (80),  and  join  PH.     PH  is  ±  to  AB. 

Since  D  and  E  are  points  in  the  circumf.,  CFG,    (Const.) 

P  is  equidistant  from  D  and  E ; 

also  H  is  equidistant  from  D  and  E, 

.'.  PH  is  ±  to  DE  at  its  mid  point  H, 

i.e.,  PH  is  drawn  A- to  AB. 

93.  Cor.  The  perpendicular  is  the  short- 
est line  that  can  he  drawn  to  a  given  line 
from  a  given  point  without  it. 

For  if  PA  is  a  perpendicular,  and  PB  a 
line  oblique  to  MN,  then  Z.  B  <,  right  Z  A 
(85)  ;  hence  PA<,PB  (86). 


*  That  is,  the  given  line  may  be  produced  to  any  extent,  if  necessary. 


PERPENDIC  ULA  RS.  47 

94.  Definition.  The  distance  of  a  point  from  a  line  is 
the  distance  from  the  point  to  the  foot  of  the  perpendicu- 
lar to  the  line.  Thus  the  distance  of  P  from  MN  is  the 
perpendicular  distance  PAy  this  being  the  least  distance 
from  P  to  any  point  in  MN  (93). 


Proposition  XXII.     Problem. 

95.  To  draw  a  perpendicular  to  a  given  line  at  a 
given  point  in  it. 


Given :  A  point  P  in  a  straight  line  AB  ; 

Required :  To  draw  a  perpendicular  to  ^i?  at  P. 

On  AB  lay  off,  on  each  side  of  P,  equal  parts  PC,  PD. 

Find  a  point  E  that  is  equidistant  from  c  and  D.     (78) 

t      Join  EP  ;  then  EP  is  ±  to  ^IP  at  P. 

Since  P  is  equidistant  from  C  and  D,  ) 

.       ....  .  .       [      (Const.) 

and  E  IS  equidistant  from  C  and  D,  ) 

EP  is  ±  to  CD  at  its  mid  point  P,  (75) 

i.e.,  EP  is  drawn  _L  to  AB  at  P.  q.e.f. 

Scholium.  In  this  problem  we  have  what  may  be  regarded 
as  a  special  case  of  the  bisection  of  an  angle.  What  is  re- 
quired is,  in  fact,  to  bisect  a  straight  angle  (compare  81). 

Exercise  53.  At  a  given  point  in  a  line  make  an  angle  equal  to  half 
a  right  angle. 

64.  From  a  given  point  without  a  line  draw  a  line  making  with  the 
given  line  an  angle  equal  to  half  a  right  angle. 


48  PLANE   GEOMETRY.  — BOOK  I. 

Proposition  XXIIT.     Theorem. 

96.  Any  point  in  the  perpendicular  at  the  mid 
point  of  a  line  is  equidistant  from  the  extremities 
of  the  line. 


Given :  Any  point  P  in  CD,  the  _L  at  the  mid  point  of  AB  ; 
To  Prove :      P  is  equidistant  from  A  and  B. 

Join  P  with  A  and  B. 
Then  since  DA  =  DB  (Hyp.),  and  PD  is  common, 

also  rt.  Z  PDA  =  rt.  Z  PDB,  (42) 

A  PDA  =  A  PDB,  (66) 

.-.  PA  =  PB,  (70) 

i.e.,  P  is  equidistant  from  A  and  B.  q.e.d. 

97.  Cor.  1.  Conversely,  cuiy  point  that  is  equidistant  from 
the  extremities  of  a  line  will  lie  in  the  perpendicular  through 
its  mid  point. 

For  the  line  joining  that  point  with  the  mid  point  of  the 
line  is  perpendicular  to  the  latter  at  its  mid  point  (75). 

98.  Cor.  2.  Aiiy  point  not  in  the  perpendicular  through 
the  midpoint  of  a  line  is  unequally  distant  from  its  extremities. 

For  if  it  were  equidistant  from  those  extremities,  it 
would  be  in  that  perpendicular  (97). 

Exercise  55.  In  the  diagram  for  Prop.  XXIII.,  show  that  angle 
CPA  =  angle  CPB. 

56.  In  the  same  diagram,  what  condition  must  be  fulfilled  in  order 
that  triangle  PAB  may  be  equiangular  ? 


PERPENDICULARS.  49 

Proposition  XXIV.     Theorem. 

99.  If  from  any  point  in  a  perpendicular  to  a 
given  line  different  oblique  lines  he  drawn  to  the 
given  line,  then 

1°.  Oblique  lines  drawn  to  equal  distances  from 
the  foot  of  the  perpendicular  are  equal. 

2°.  Of  oblique  lines  drawn  to  unequal  distances 
from  the  foot  of  the  perpendicular,  the  more  remote 
is  the  greater. 


Given :  PD  ±  to  MN,  and  PA,  PB,  PC,  drawn  oblique  to  MN, 
so  that  AD  is  equal  to  BD,  but  CD  is  greater  than  AD  ; 

To  Prove :  PA  is  equal  to  PB,  but  PC  is  greater  than  PA  oi PB. 

1°.      Since  P  is  in  the  ±  at  the  mid  point  of  AB,    (Hyp.) 
PA  =  PB.  Q.E.D.      (96) 

2°.               Since  exterior  Zpac>  right  Z  D,  (83) 

but  right  Zd>Zc,  (85) 

ZpaoZc,  (a.f.) 

.-.  PC  >  PA  or  its  equal  PB.       q.e.d.  (86) 

100.  Cor.  Only  two  equal  straight  lines  can  be  drawn 
from  a  point  to  a  given  straight  line,  one  on  each  side  of  the 
perpendicular  from  that  point. 

Exercise  57.  From  a  point  without  a  line,  show  that  only  two 
oblique  lines  can  be  drawn  so  as  to  make  equal  angles  with  the  given 
line. 

58.  If  oblique  lines  drawn  from  a  given  point  without  it  make 
unequal  interior  angles  with  a  line,  that  which  makes  the  lesser  angle 
is  the  greater. 

Geom.  — 4 


50  PLANE   GEOMETRY.  — BOOK  I. 

Proposition  XXV.     Theorem. 

101.   Any  point  in  the  bisector  of  an  angle  is  equi- 
distant from  its  sides;  and  conversely. 


1°.    Given :  A  point  P  in  the  bisector  of  angle  BAC; 
To  Prove :   P  is  equidistant  from  AB  and  AC. 

Draw  PD  ±  to  AB  and  PE  _L  to  AC.     Then 
since  AP  is  the  hypotenuse  of  rt.  A  PDA,  PEA,   (Const.) 
and  Z  PAT)  =  Z  PAE,  (Hyp.) 

right  A  PAD  =  right  A  PAE,  (73) 

.-.    PD  =  PE.  Q.E.D.       (70) 

2°.  Given :  In  angle  BAC,si  point  P  equidistant  from  AB  and  AC  ; 
To  Prove:  The  line  joining  PA  bisects  angle  BAC. 

Draw  PD  ±  to  AB,  PE  J-  to  AC,  and  join  PA. 
Then  since  PD  =  PE  (Hyp.),  and  AP  is  common, 

also  right  Z  D  =  right  Z  E,  (Const.) 

right  A  PAD  =  right  A  PAE,  (72) 

.'.  Zpad  =  Zpae,  (70) 

.-.  PJ[  bisects  Z^^e.  q.e.d. 

Exercise  59.  Show  that  Prop.  XXIII.  is  a  particular  case  of  XXV. 

60.  Prove  that  any  point  not  in  the  bisector  of  an  angle  is  unequally 
distant  from  its  sides. 

61.  Enunciate  and  prove  the  converse  of  Exercise  60. 

62.  Prove  that  a  perpendicular  to  the  bisector  of  an  angle  makes 
equal  angles  with  the  sides. 


PARALLELS.  51 

PARALLELS. 

102.  Parallel  lines  are  such  as  lie  in  the  same  plane,  but 

cannot  meet,  however  far  pro- 
duced in  either  direction. 

103.  A  transversal  is  a 
straight  line  that  is  trans- 
verse to,  that  is,  meets  or  in- 
tersects, a  set  of  two  or  more 
straight  lines. 

104.  When  a  transversal  EF  intersects  two  lines  AB,  CD, 
then 

{a)  The  four  A  3,  4,  5,  6,  within  AB,  CD,  are  interior  A. 

(b)  The  four  A  1,  2,  7,  8,  without  AB  and  CD  are  exterior  A. 

(c)  The  pairs  of  interior  A,  3-5,  4-6,  situated  on  opposite 
sides  of  the  transversal,  are  alternate-interior  A. 

(d)  The  pairs  of  exterior  A,  1-7,  2-8,  situated  on  opposite 
sides  of  the  transversal,  are  alternate- exterior  A. 

(e)  The  pairs  of  A,  1-5,  4-8,  2-6,  3-7,  situated  on  the  same 
side  of  the  transversal,  but  one  exterior,  the  other  interior, 
are  corresponding  A. 

Angles  hereafter  referred  to  simply  as  alternate  are  to  be 
understood  as  being  alternate-interior,  those  most  frequently 
mentioned. 

E 

105.  Axiom  10.  Tivo  inter-  a  -^ — ■— ,__^^ — - 
secting  lines  cannot  both  be  par-  f 
allel  to  a  third  line. 

C  D 

Thus,  if  AB  is  parallel  to 
CD,  EF  cannot  be  so  ;  and  vice  versa. 

Exercise  63.   Enunciate  and  prove  the  converse  of  Exercise  62. 

64.  Prove  that  the  bisector  of  an  angle  of  a  triangle  divides  the 
opposite  side  equally  or  unequally  according  as  the  other  sides  are  or 
are  not  equal. 


62 


PLANE   GEOMETRY.  — BOOK  I. 


Proposition  XXYI.     Theorem. 

106.  If  a  transversal  is  perpendicular  to  two  lines 
in  the  same  plane,  these  lines  are  parallel. 


A 

E                                               B 

C 

F                                              D 

Given :        EF  perpendicular  to  AB  and  also  to  CD  ; 
To  Prove :  AB  is  parallel  to  CD. 

Since  AB  and  CD  are  each  ±  to  EP,         (Hyp.) 
AB  cannot  meet  CD  in  any  point  on  either  side  of  EF,  no 
matter  how  far  produced, 
since  from  the  same  point  there  cannot  be  two  Js  to  a  line,  (51) 

.-.    AB  is  II  to  CD.  Q.E.D.       (102) 


Proposition  XXVII.     Theorem. 

107.    If  a  transversal  is  perpendicular  to  one  of 
two  parallels,  it  is  perpendicular  to  the  other  also. 


E 


Given :  Two  parallels  AB,  CD,  and  ^C  a  transversal  ±  to  AB ; 
To  Prove:         AC  i^  perpendicular  to  CD. 


PARALLELS. 


At  C  suppose  CE  drawn  1.  to  AC. 
Then  since  CE  is  II  to  AB, 


63 

(95) 
(106) 
(Hyp.) 


and  CD  is  II  to  AB, 

CD  must  coincide  with  CE, 

since  otherwise  there  would  be  through  C  two  intersecting 
parallels  to  the  same  line,  which  is  impossible.    (Ax.  10) 

.-.    CD  \^  A.  to  AC.  Q.E.D. 

108.  CoR.  If  AB,  CD,  are  each 
parallel  to  EF,  then  AB  is  parallel 
to  CD. 

For  a  ±  to  EF  must  be  ±  to 
both  AB  and  CD  (107). 


A 

B 

C 

D 

E 

F 

Proposition  XXVIII.     Problem. 

109.    Through   a  given    point    without   a    line  to 
draw  a  parallel  to  the  line. 


Given :        A  point  P  without  a  given  line  AB  ; 
Required:  To  draw  through  P  a  hne  parallel  to  AB. 

From  P  draw  PC  ±  to  AB,  or  AB  produced,       (92) 
and  from  P  draw  PD  A.  to  PC.  (95) 

PD  is  the  parallel  required. 
For  since  PC  is  _L  to  AB,  and  also  to  PD,  (Const.) 

.-.   PD  is  II  to  AB.  Q.E.F.       (106) 


54 


PLANE    GEOMETRY.  — BOOK  I. 


Proposition  XXIX.     Theorem. 

110.  Two  lines  are  parallel  if  a  transversal  to  those 
lines  makes  the  alternate  angles  equal;  and  conversely. 


A 

B 

p 

R^ 

y^ 

^ 

c 

y^F 

Q 

T> 

1°.  Given:  EF  transverse  to  AB,  CD,  making  the  alternate 
angles  E,  F,  equal ; 

To  Prove:  AB  i^  parallel  to  CD. 

Find  O,  the  mid  point  of  EF  (80),  draw  OP  ±  to  AB  (92), 
and  produce  PO  to  meet  CD  in  Q. 

Then  since  /.F  =  Z.E  (Hyp.),  and  /.FOQ  =  /.  EOF,    (50) 
and  OF  =  OE,  (Const.) 

AOQF  =  AOPE,  (63) 

.-.  Zq  =  rt.  Z  P,  (70) 

.-.    AB  is  II  to  CD.  Q.E.D.       (106) 

2°.  Given:  A  transversal  EF  meeting  the  parallels  AB,  CD, 
in  E,  F,  respectively ; 

To  Prove :   Angle  E  is  equal  to  alternate  angle  F. 

Making  the  same  constructions  as  in  1°, 
Since  AB  is  II  to  CD  (Hyp.),  and  PQ  is  _L  to  AB, 
PQ  is  also  ±  to  CD  (107),  and  Q  is  a  rt.  Z  ; 
also  Z  EOF  =  Zfoq  (50),  and  OE  =  OF, 
.:  rt.  A  OPE  =  rt.  AOQF, 


(Const.) 


(Const.) 
(73) 


.-.  Ze  =  Z  F. 

111.    Scholium.    It  is  sufficient 
to  prove  the  foregoing  propositions,     ~ 
as  above,  for  one  pair  of  alternate 
angles  only ;  since  the  equality  of 
any  one  pair  of  the  alternate  angles 


Q.E.D.      (70) 


jzi. 


8^7 


PARALLELS.  55 

.formed  by  a  transversal  and  two  lines,  entails  the  equality 
of  every  pair.  For  if  Z  4  =  Z  6,  then  Z  2  =  Z  8  (50)  ;  also 
Z3  =  Z5,  andZl=  Z7  (44). 


Proposition  XXX.     Theorem. 

112.  Two  lines  are  parallel  if  a  transversal  to 
those  lines  makes  the  corresponding  angles  equal; 
and  conversely. 


T 


1°.  Given :  EF  transverse  to  AB,  CD,  making  the  corresponding 
angles  F,  g,  equal ; 

To  Prove :  AB  h  parallel  to  CD. 

Since  Af  =  /.g  (Hyp.),  and  Z  ^  =  Z  G,  (50) 

/.E  =  /.F,  (Ax.  1) 

.-.    AB  is  II  to  CD.  Q.E.D.       (110')* 

2°.    Given:  A  transversal  EF  meeting  the  parallels  AB,  CD, 
in  E,  F,  respectively ; 

To  Prove :  Angle  F  is  equal  to  the  corresponding  angle  G. 

Since  EF  is  transverse  to  the  II 's  AB,  CD,     (Hyp.) 

Ze  =  Zf.  (110") 

But  Z  J5:  =  Z  G,  (50) 

.\Z.F  =  /.G.  Q.E.D.        (Ax.  1) 

*  When,  as  here,  reference  is  made  to  a  proposition  consisting  of  two 
parts,  these  parts  will  be  distinguished  by  accents;  thus,  110'  and  110". 


56  PLANE   GEOMETRY.  — BOOK  I. 

Proposition  XXXI.     Theorem. 

113.  Two  lines  are  parallel  if  a  transversal  to 
those  lines  mahes  the  interior  angles  on  the  same 
side  supplementary;  and  conversely. 


A 

yo 

B 

/^' 

C 

yF 

D 

X 


1°.  Given:  EF  transverse  to  AB,  CD,  making  the  interior 
angles  E,  F,  supplementary  ; 

To  Prove :  AB  is  parallel  to  CD. 

Since  Z  F  is  the  supplement  of  Z  E,         (Hyp.) 

and  Z  G  is  the  supplement  of  Z  E,  (46) 

Z  F  =  the  corresponding  Z  G,  (Ax.  1) 

.-.   AB  is  II  to  CD.  Q.E.D.      (112') 

2°.  Given :  A  transversal  EF  meeting  the  parallels  AB,  CD,  in 
E,  F,  respectively ; 

To  Prove:  The  interior  angles  E  and  F  are  supplementary. 

Since  EF  is  transverse  to  the  ll's  AB,  CD,    (Hyp.) 

Z  F  =  corresponding  Z  G.  (112") 

But  Z  jE;  is  the  supplement  of  Z  G,  (46) 

.*.  Z  ^  is  the  supplement  of  Z  F.     q.e.d.     (44) 

114.  Cor.  If  a  transversal  to  two  lines  makes  the  sum  of 
the  interior  angles  on  one  side  less  than  a  straight  angle,  these 
lines  will  meet  if  produced  on  that  side. 

For  if  they  did  not  meet,  they  would  be  parallel,  and 
those  interior  angles  would  be  supplementary  (113")  ;  that 
they  must  meet  on  that  side  of  the  transversal  on  which  the 
interior  A  are  less  than  a  straight  Z  follows  from  Art.  84. 


PARALLELS.  57 

115.  Definition.  Parallel  lines  are  said  to  lie  in  the 
same  or  opposite  directions,  according  as  they  are  on  the 
same  or  opposite  sides  of  the  transversal  through  the  points 
from  which  they  are  supposed  to  be  drawn. 


Proposition  XXXII.     Theorem. 

116.  Angles  whose  corresponding  sides  are  parallel 
are  equal  or  supplementary. 


Given:  AB  parallel  to  DE,  and  AC  parallel  to  HF,  forming 
angles  at  A  and  D  ; 

To  Prove :  Angle  A  is  equal  to  angle  D,  and  angle  A  is  supple- 
ment of  angle  EI)H. 

Let  AC,  DE,  produced  if  necessary,  meet  in  G. 
Since  AB  is  II  to  DE,  and  ^C  is  transverse  to  both, 

Ag  =  Aa;  (112") 

Since  AG  is  II  to  DF,  and  DE  is  transverse  to  both, 

Z.G  =  Z.D,  (112") 

.-.  /.A  — Ad;  q.e.d.     (Ax.  1) 

also  Z  D  is  supp.  of  Z  EDH,  (46) 

.-.  Z  ^  is  supp.  of  Z  ^DiT.         q.e.d.     (44) 

117.  Scholium.  The  angles  are  equal  if  both  corre- 
sponding pairs  of  sides  lie  in  the  same  or  opi^osite  direc- 
tions; they  are  supplementary  if  one  pair  has  the  same  and 
the  other  opposite  directions. 


58 


PLANE   GEOMETRY.  — BOOK  I. 


Proposition  XXXIII.     Theorem. 

118.  Angles  whose  corresponding  sides  are  perpen- 
dicular to  each  other  are  either  equal  or  supple- 
mentary. 

F 


IK 


sB 


Given:  AB  perpendicular  to  DE  and  AC  perpendicular  to  DF, 
forming  angles  at  A  and  D  ; 

To  Prove :  Angle  A  is  equal  to  angle  D,  and  angle  A  is  the 
supplement  of  angle  EDG. 

At  A  draw  AH  1.  to  AB,  and  AK  1.  to  AC.         (95) 
Since  BAH  and  CAK  are  rt.  A,  (Const.) 

Z.A  =  /.  HAK,  (44) 

(each  being  the  comp.  of  Z  bak.) 
Since  AB  is  ±  DE  and  AH,  and  ^C  is  _L  to  I)F  and  AK, 

(Hyp.  and  Const.) 
DE  is  II  to  AH,  and  DF  is  ||  to  AK ;  (106) 

and  they  lie  in  the  same  directions, 

.-.  Zd  =  Zhak,  (117) 

.*.  Za  =  Zd;  q.e.d.     (Ax.  1) 

also  Zd  is  supp.  of  Z  EDG,  (46) 

.*.  Z^  is  supp.  of  Zedg.        q.e.d.     (44) 

119.  Scholium.  The  angles  are  equal  if  both  are  acute 
or  both  obtuse. 


Exercise  65.    In  the  diagram  for  Prop.  XXIX.,  show  that  PF  being 
joined,  a  line  drawn  through  Q  parallel  to  PF  will  pass  through  E. 


PARALLELS.  59 

Proposition  XXXIV.     Theorem. 

120.  The   sum  of  the   angles    of   any   triangle    is 
equal  to  a  straight  angle. 


Given:  Any  triangle  ABC ; 

To  Prove :  Za-\-Zb-\-Zc[&  equal  to  a  straight  angle. 

Through  A  draw  DAE  II  to  BC.  (109) 

•.•  DE  is  II  to  BC,  and  AB  is  transverse  to  both,   (Const.) 
Zb  =  Z1;  (110") 

•••  DE  is  II  to  BC,  and  ^C  is  transverse  to  both, 

ZC  =  Z2,  (110") 

.'.  Za  +  Zb  +  Zc  =  Za-{-Z1-{-Z2,      (Ax.2) 

i.e.,  Za  +  Zb  ■\-Zc  =  Z  DAE,  a  st.  Z.     q.e.d.     (47) 

121.  Cor.  1.  Each  angle  of  a  triangle  is  the  supplement 
of  the  sum  of  the  other  two. 

122.  Cor.  2.  An  exterior  angle  of  a  triangle  is  equal  to 
the  sum  of  the  two  remote  interior  angles. 

For  its  supplement  is  also  the  supplement  of  those  two. 

123.  Cor.  3.  The  acute  angles  of  a  right  triangle  are  com- 
plementary. 

124.  Definition.  A  plane  polygon  is  a  portion  of  a  plane 
bounded  by  straight  lines.  Its  sides,  angles,  and  vertices  are 
defined  as  for  the  triangle,  which  is  a  three-sided  polygon. 
The  polygons  treated  of  are  to  be  understood  as  being 
convex;  i.e.,  such  that  no  side,  if  produced,  will  fall  within 
the  polygon. 


60  PLANE    GEOMETRY.  — BOOK  I. 

Proposition  XXXV.     Theorem. 

125.  The  Sinn  of  the  interior  angles  of  a  polygon  is 
equal  to  as  many  straight  angles  as  the  figure  has 
sides,  less  two. 


E 


D 


Given :        A  polygon  AB  CDEF  •  •  •  of  7i  sides ; 

To  Prove  :Za+Zb-{-Zc-\ =  {n  —  2)  straight  a; 

From  any  vertex,  as  A,  draw  lines  to  C,  D,  E,  •••. 

Since,  except  the  two  that  meet  at  A,  each  of  the  n  sides 
of  the  polygon  is  the  base  of  a  triangle  having  its 
vertex  at  A,  (Const.) 

the  number  of  triangles  =  ?i  —  2 ; 

.*.  the  sum  of  the  int.  A  of  the  A=  (n  —  2)  st.  A,     (120) 

•.•  the  sum  of  the  int.  A  of  the  polygon  =  {n  —  2)  st.  A.  q.e.d. 

126.  Cor.  1.    If  the  sides  of  a  polygon  be  produced,  going 
round  in  the  same  order,  the  sum   of  the 

exterior  A  thus  formed  =  two  straight  A. 

For  each  interior  Z  A  with  its  exterior 

Z  a  is  equal  to  a  straight  Z  (46) .     Hence       /A ^Ml 

the  sum  of  all  the  angles,  interior  and 

exterior,  =  n  straight  A.     But  the  interior  A  alone  =  (n  — 2) 

straight  A ;  hence  the  exterior  A  =  two  straight  A. 

127.  Cor.  2.   Each  interior  angle  of  an  equiangular  poly- 
gon is  equal  to  straight  angles. 


QUADRILA  TERALS, 


61 


QUADRILATERALS. 

128.  A  quadrilateral  is  a  polygon  bounded  by  four  sides. 

129.  A  trapezium  is  a  quadrilateral  having  no  two  of  its 
sides  parallel. 

130.  A  trapezoid  is  a  quadrilateral  having  two  of  its  sides 
parallel. 

131.  A  parallelogram  is  a  quadrilateral  having  its  oppo- 
site sides  parallel. 


PARALLELOGRAM 


132.  A  rectangle  is  a  right-angled  parallelogram. 

133.  A  square  is  an  equilateral  rectangle. 

134.  A  rhombus  is  a  parallelogram  which  is  equilateral 
but  not  equiangular. 

135.  A  diagonal  is  a  line  joining  any  two  nonadjacent 
vertices  of  any  polygon. 


RECTANGLE 


SQUARE 


RHOMBUS 


Exercise  66.   Prove  that  each  angle  of  an  equilateral  triangle  is 
one  third  of  a  straight  angle. 

67.  Prove  that  if  an  angle  at  the  base  of  an  isosceles  triangle  is  one 
half  of  the  vertical  angle,  the  latter  is  a  right  angle. 

68.  If  an  angle  at  the  base  of  an  isosceles  triangle  is  n  times  the 
vertical  angle,  what  fraction  is  the  latter  of  a  straight  angle  ? 


62  PLANE   GEOMETRY.  —  BOOK  I. 

Proposition  XXXVI.     Theorem. 

136.  The  opposite  sides  and  angles  of  a  parallelogram 
are  equal. 

D  c 


Given :  A  parallelogram  AB CD,  oi  AC; 

To  Prove :  AB  =  CD,  AD  =  BC;  A  A  =Z  C,  and  Z  B  =Z  D. 

Since  AB  is  li  to  CD,  and  BC\^  II  to  AD,      (Hyp.) 
and  the  corresponding  lines  lie  in  opposite  directions,   (115) 
Z^  =  Z  c,  and  Z5  =Zi).  q.e.d.    (117) 

Draw  the  diagonal  DB. 

Then  since  BD  is  transverse  to  the  ll's  AB,  CD, 

Abdc  =  Aabd;  (110") 

similarly,    Z  CBD  =  Z  ADB,  and  BD  is  common, 

.:  AABD  =  ACBD,  (63) 

.-.  AB  =  CD,  and  AD  =  BC.       Q.E.D.       (70) 

137.  Definition.  An  intercept  is  the  straight  line,  or 
part  of  a  line,  intercepted  between  two  other  lines.  Thus  a 
diagonal  is  an  intercept  through  the  angular  points. 

138.  Cor.  1.  Parallel  interceiyts  between  parallels  are  equal. 

139.  Cor.  2.    Two  parallels  are  everywhere  equally  distant. 

140.  Cor.  3.  A  diagonal  divides  a  parallelogram  into  two 
equal  triangles. 

This  was  proved  in  the  demonstration  of  Prop.  XXXVI. 


QUADRILA  TERALS.  63 


Proposition  XXXVII.     Theorem. 

141.  //  the  opposite   sides   or   angles  of  a  quadri- 
lateral are  equal,  the  figure  is  a  parallelogram. 


Given:  In  the  quadrilateral  ABCD, 

1°,  AB  equal  to  CD,  EC  equal  to  AD ; 

2°,  angle  A  equal  to  angle  c,  angle  B  to  angle  D  ; 
To  Prove :  In  either  case,  AB  CD  is  a  parallelogram. 

Draw  the  diagonal  BD.     Then 

1°.    Since  AB  =  CD,  AD  =  BC,  and  BD  is  common,  (Hyp.) 

A  ABD  =  A  CBD,  (69) 

.-.  Zabd  =  Zbdc,  and  Zadb=Zcbd,        (70) 

.♦.  AB  is  11  to  CD,  and  AD  is  II  to  BC,  (HO') 

.-.  ABCD  is  a  parallelogram,      q.e.d.     (131) 

2°.  .            Since  Za  =  Zc,  and  Zb  =Zd,  (Hyp-) 

Za-\-Zb  =  Zc-^Zd.  ■  (Ax.  2) 

But  Z  ^  +  Z  5  -f  Z  (7  -f  ^  i>  =  4  rt.  Zs,  (125) 

.-.  Z^+ZJ5  =  2  rt.  Zs,  (Ax.  7) 

.-.     ^DislltoJSC;  (113') 

similarly,  AB  is  11  to  CD, 

.-.  ABCD  is  a  parallelogram,     q.e.d.     (131) 

Exercise  69.  Show  that  an  angle  of  a  triangle  is  obtuse  or  acute, 
according  as  it  is  greater  or  less  than  the  sum  of  the  other  angles. 

70.  Show  that  the  bisectors  of  opposite  angles  of  a  parallelogram 
are  parallel. 


64  PLANE   GEOMETRY.  — BOOK  I. 

Proposition  XXXVIII.     Theorem. 

142.  If  two  sides  of  a  quadrilateral  are  parallel 
and  equal,  the  figure  is  a  parallelogram. 

D  c 


Given:  In  the  quadrilateral  ABCD,  AB  equal  and  parallel  to 
CD; 

To  Prove:  ABCD  is  a  parallelogram. 

Draw  the  diagonal  BD. 
Then  since  5Z>  is  transverse  to  the  parallels  AB,  CD, 

A  ABB  =  alt.  Z  CDB  ;  (110") 

also  AB  =  CD,  and  BD  is  common,  (Hyp.) 

.'.   AABD  z=ACBD,  (66) 

.♦.   Z  ADB  =  Z  CBD,  (70) 

.-.   AD  is  II  to  BC,  (110') 

.-.  ABCD  is  a  parallelogram,      q.e.d.    (131) 

Exercise  71.  A  quadrilateral  whose  diagonals  bisect  each  other  is 
a  parallelogram. 

72.  In  a  quadrilateral  ABCD,  if  angle  A  is  supplementary  to  angle 
B,  and  angle  B  to  angle  C,  then  the  figure  is  a  parallelogram. 

73.  If  two  parallelograms  have  an  angle  of  the  one  equal  to  an 
angle  of  the  other,  they  are  mutually  equiangular. 

74.  A  parallelogram  whose  diagonals  are  equal  is  a  rectangle. 

75.  A  parallelogram  whose  diagonals  bisect  the  angles  through 
whose  vertices  they  pass  is  equilateral. 

76.  If  one  angle  of  a  parallelogram  is  a  right  angle,  all  its  angles 
are  right  angles. 


Q  UA  DRILA  TERA  LS.  Q^ 

Proposition  XXXIX.     Theokem. 

143.  Tivo  parallelogj^aiivs  are  equal  if  two  sides  and 
the  included  angle  of  the  one  are  respectively  equal 
to  two  sides  and  the  included  angle  of  the  other. 


A' 

Given:  In  parallelograms  AC  and  A^c\  AB  equal  to  A^B\  AD 
equal  ta  A^D\  and  angle  A  equal  to  angle  A^ ; 

To  Prove :     Parallelogram  ^ C  is  equal  to  parallelogram  A'c'. 

Place  AC  upon  A'c',  so  that  Za^Za'. 

Since  AB  =  a'b',  and  AD  =  a'd',  (Hyp.) 

B  will  fall  on  B',  and  D  on  7)'.  (14) 

•.•  DC,  D'C'  are  II  to  the  coinciding  lines  AB,  a'b',  (Hyp.) 

DC  must  take  the  direction  of  D'c' ;  ) 

>  (Ax.  10) 
similarly,  BC  must  take  the  direction  oi  B'c^,  ) 

.'.  C,  the  common  point  of  DC,  BC,  must  fall  on  c',  the 

common  point  of  D'c',  B'c', 

.-.  parallelogram  ^C  =  parallelogram  ^'C'.     q.e.d.     (61) 

144.  Scholium.  Hence  two  adjacent  sides  and  the  in- 
cluded angle  are  said  to  determine  a  parallelogram. 

Exercise  77.  If  both  diagonals  of  a  parallelogram  are  drawn,  of 
the  four  triangles  thus  formed,  those  opposite  are  equal. 

78.  The  bisectors  of  those  angles  of  a  parallelogram  that  have  a 
side  common,  meet  at  right  angles. 

79.  In  the  diagram  for  Prop.  IV.,  if  OA  be  made  equal  to  OB,  and 
OC  to  OD,  and  AC,  AD,  BC,  BD,  be  drawn,  ABCD'^flW  be  a  paral- 
lelogram. What  condition  must  be  fulfilled  so  that  the  figure  will  be 
equilateral  ? 

Geom.  — 6 


66  PLANE   GEOMETRY.— BOOK  I. 

Proposition  XL.     Theorem. 

145.  An  intercept  passing  through  the  mid  point 
of  a  diagonal  of  a  parallelogram  is  bisected  in  that 
point,  and  cuts  off  equal  parts  on  the  intercepting 
sides, 

o 


A        E 


Given :  In  parallelogram  A  c,  an  intercept  EF  tlirougli  0,  the 
mid  point  of  diagonal  BD  ; 

To  Prove :  OE  is  equal  to  OF,  and  EB  to  FD. 

Since  OB  =  OD,  (Hyp.) 

also  Z  BDF  =  Z  DBF  (110"),  and  ^  at  0  are  equal,      (50) 

Aboe  =  Adof,  (63) 

.-.    OE—  OF,  and  EB  =  FD.  Q.E.D.      (70) 

146.  Cor.  The  diagonals  of  a  parallelogram  bisect  each 
other. 

Exercise  80.  In  the  diagram  for  Prop.  XV.,  if  A  and  B  be  each 
joined  with  C  and  D,  then  ADBC  will  be  a  square  or  a  rhombus 
according  as  EA  is  or  is  not  equal  to  EC. 

81.  If  two  equal  isosceles  triangles  are  constructed  on  opposite 
sides  of  the  same  base,  what  sort  of  a  figure  is  obtained  ? 

82.  Each  interior  angle  of  an  equiangular  polygon  of  six  sides  is 
the  double  of  each  interior  angle  of  an  equilateral  triangle. 

83.  How  many  sides  has  an  equiangular  polygon  if  an  interior  is 
equal  to  one  half  of  an  exterior  angle  ? 

84.  How  many  sides  has  an  equiangular  polygon  if  an  interior  is 
the  double  of  an  exterior  angle  ? 

85.  In  the  diagram  for  Prop.  XL.,  show  that  DEAD  =  OFCB. 


'  QUADBJLATERALS.  67 

Proposition  XLI.     Theorem. 

147.   An  intercept  parallel  to  the  base  of  a  triangle 
and  bisecting  one  side  bisects  the  other  also. 


Given:  In  triangle  ABC,  an  intercept  OE  parallel  Xo  AB  and 
bisecting  BC  m  0 ; 

To  Prove :  ^  C  is  bisected  in  E. 

Complete    the    parallelogram   AD   by   drawing  BD,    CD, 
parallel  to  AC,  AB,  respectively,  and  produce  EO  to 
meet  BD  in  F. 
Since  EF  is  an  intercept  through  the  mid  point  oiBC,  (Hyp.) 

BF  =  EC.  (145) 

But  BF  =  AE,  (136) 

(since  AF  is  a  parallelogram  by  construction,) 

.'.AE=EC.  Q.E.D.       (Ax.  1) 

148.  Cor.  Conversely,  if  OE  bisects  both  AC  and  BC,  then 
OE  is  parallel  to  AB,  and  OE  is  equal  to  i  AB. 

For  OE  must  coincide  with  the  parallel  to  ^^  through  0, 
since  that  parallel  must  pass  through  E,  the  mid  point  of 
BC  (147).  Also  OE  is  equal  to  ^fe  (145),  and  FE  is  equal 
to  AB  (136). 

Exercise  86..  In  the  diagram  for  Prop.  XLI.,  show  that  OEAB 
can  be  superposed  on  OFDC. 

87.  In  the  diagram  for  Prop.  XLII.,  if  BD,  CG,  be  drawn,  show 
that  these  lines  will  intersect  in  the  mid  point  of  FH. 

88.  In  the  same  diagram,  if  AD  =  BC,  then  the  angles  A  and  B  are 
equal. 


68  PLANE   GEOMETRY.  — BOOK  I. 

Proposition  XLTI.     Theorem. 

149.   An  intercept  parallel  to  the  bases  of  a  trape- 
zoid and  bisecting  one  of  the  nonparallel  sides  bisects 

the  other  also. 

D  c 

H 


Given:  In  trapezoid  AC,  EF  parallel  to  AB,  bisecting  AD  in 
E  and  meeting  BC  in  F ; 

To  Prove :  ^C  is  bisected  in  F. 

Draw  DG  II  to  BC,  meeting  AB,  EF,  in  G,  H,  resp. 

Since  in  A  DAG,  EH  is  II  to  AG  and  bisects  AD,    (Hyp.) 

DG  i^  also  bisected  in  //.  (147) 

But  BH,  HC,  are  parallelograms,  (Const.) 

.-.  BF  =  GH,  and  FC  =  HD  or  GH,  (136) 

.:  BF  =  FC.  Q.E.D.      (Ax.  1) 

150.  Cor.  Conversely,  if  EF  bisects  both  AD  and  BC,  then 
EF  is  parallel  to  AB  and  equal  to  ^  (AB  -{-CD). 

For  EF  must  coincide  with  the  parallel  to  ^J5  through  E, 
since  that  parallel  must  pass  through  F,  the  mid  point  of 
BC  (149).  Also  EF  =  EH  -\-  HF  =  i  AG  -h  i  {BG  +  CD) 
=  i{AB-\-CD). 


Exercise  89.  In  the  diagram  for  Prop.  XLII.,  if  CK  be  drawn 
parallel  to  AD,  then  will  BK  be  equal  to  AG,  and  GK=2DG  -  AB. 

90.  If  through  P,  the  mid  point  of  AE,  FQ  be  drawn  parallel  to  AB 
to  meet  BF  in  Q,  then  will  ^^  be  one  fourth  of  ^C,  and  PQ  wiU 
be  equal  to  ^  (3AB+  CD). 


EXERCISES.  69 

Proposition  XLIII.     Theorem. 

151.  If  a  series  of  parallels  wiahe  equal  intercepts 
on  one  transversal,  they  make  equal  intercepts  on 
any  transversal. 


Given :  Transversals  AE,  ae,  cut  by  a  series  of  parallels  Aa, 
Bh,  Cc,  Dd,  Ee,  so  that,  the  intercepts  AB,  BC,  CD,  BE,  on  AE 
are  equal ; 

To  Prove :  The  intercepts  ab,  be,  cd,  de,  on  ae,  are  also  equal. 

If  AE  is  II  to  ae,  the  opposite  intercepts  are  equal.     (136) 

If  AE  is  not  II  to  ae,  then  in  the  trapezoid  ACca, 

since  Bb  is  II  to  the  bases  Aa,  Cc,  and  bisects  AC,  (Hyp.) 

ab  =  be.  (149) 

In  the  same  way  may  be  proved  that  be  =  cd,  and  cd  =  de. 

.-.  ab  =  bc  =  cd  =  de.        q.e.d.     (Ax.  1) 

152.  Cor.  Conversely,  if  Aa,  Bb,  etc.,  make  equal  inter- 
cepts on  AE  and  also  on  ae,  then  Aa  is  II  to  Bb,  Cc,  etc.     (loO) 


EXERCISES. 

QUESTIONS. 


91 .  Would  a  triangle  constructed  of  rods  hinged  at  their  extremities 
be  rigid ;  that  is,  incapable  of  change  of  form  ?  Would  a  parallelo- 
gram similarly  constructed  be  rigid  ? 

92.  Triangles  having  their  sides  severally  equal  have  their  angles 
also  severally  equal.    Is  the  converse  true  ? 

93.  An  acute  angle  being  given,  by  what  construction 
can  you  find  its  complement  ?     Its  supplement  ? 


70  PLANE   GEOMETRY.  — BOOK  I. 

94.  By  what  angle  is  the  supplement  of  an  angle  greater  than  its 
complement  ? 

95.  A  right  angle  being  90°,  how  many  degrees  are  there  in  the 

complement  of  an  angle  of  36°  ?    Of  45°?   Of  90°?  ^, -,^ 

How  many  in  their  supplements  ? 

96.  Bisect  an  obtuse  angle  (81) ;  also  a  straight 
angle.  Can  you  bisect  a  reflex  angle,  i.e.,  an  angle 
greater  than  a  straight  angle,  by  the  same  process  ? 

97.  In  Proposition  XXIII.,  can  you  prove  Cor.  2  independently? 

98.  Are  all  straight  lines  that  cannot  meet  parallel  ? 

99.  If  one  angle  of  a  triangle  is  23°,  what  is  the  sum  of  the  other 
angles  ? 

100.  If  one  angle  of  a  triangle  is  equal  to  the  sum  of  the  other  two, 
how  many  degrees  has  that  angle  ?    What  is  it  called  ? 

101.  If  the  vertical  angle  of  an  isosceles  triangle  is  50°,  how  many 
degrees  in  each  of  the  base  angles  ? 

102.  If  a  base  angle  of  an  isosceles  triangle  is  45°,  what  is  the 
vertical  angle  ? 

103.  In  an  isosceles  triangle,  if  each  base  angle  is  (1)  twice,  (2) 
three  times,  (3)  n  times,  the  vertical  angle,  how  many  degrees  in  each  ? 

104.  How  many  degrees  in  each  angle  of  an  equilateral  triangle  ? 

105.  An  acute  angle  of  a  right  triangle  is  f  of  the  other  acute 

angle.     How  many  degrees  in  each  ?     Generalize  by  putting  ^—  for  f . 

n 

106.  Arrange  the  following  terms  in  order  of  generality :  square, 
polygon,  rectangle,  quadrilateral^  parallelogram. 

107.  I  wish  to  cut  off  half  a  rectangular  field  by  a  straight  fence. 
Through  what  point  must  the  fence  pass  ? 

108.  The  base  of  a  triangle  is  fifty  feet.     How  long  is  the  line  join- 
ing the  mid  points  of  the  two  sides  ? 

109.  How  many  degrees  in  the  sum  of  the  interior  angles  of  a 
polygon  of  four  sides  ?    Of  five  ?    Of  six  ?    Oi  n  sides  ? 

110.  How  many  degrees  in  each  interior  angle  of  an  equiangular 
polygon  of  four  sides  ?     Of  five  ?    Of  six  ?    Of  ten  sides  ? 

111.  One  angle  of  a  parallelogram  is  double  the  other.     How  many 

degrees  in  each  ?     How  many,  if  one  is  —  of  the  other  ? 

n 


GEOMETRICAL   SYNTHESIS  AND  ANALYSIS.     71 

GEOMETRICAL  SYNTHESIS  AND  ANALYSIS. 

In  the  demonstration  of  propositions  we  have  usually 
proceeded  by  the  method  of  synthesis,  or  direct  proof;  that 
is,  taking  as  a  basis  certain  admitted  truths,  we  built  upon 
these  a  demonstration  of  the  truth  we  wished  to  establish. 
In  other  words,  in  synthesis  we  reason  from  admitted  prin- 
ciples to  consequences  (see,  for  example,  72).  This,  how- 
ever,^  though  usually  the  most  convenient  way  of  presenting 
the  proof  of  a  proposition,  does  not  show  how  that  proof 
was  invented;  we  are  given  a  result,  not  the  process  by 
which  the  result  was  reached. 

In  the  method  of  analysis  we  proceed  in  the  opposite 
way ;  that  is,  assuming  as  true  the  conclusion  we  wish  to 
establish,  we  reason  back  to  principles.  If  we  are  led  back 
to  principles  already  known  as  true,  we  can  take  these  as 
the  basis  of  a  synthetic  proof  of  the  conclusion  we  wish  to 
establish.  If,  on  the  other  hand,  we  are  led  to  a  contra- 
diction of  a  known  truth,  we  know  that  the  assumed  con- 
clusion was  false. 

This  indirect  method  is  often  made  use  of  in  the  demon- 
stration of  theorems  for  which  it  would  be  inconvenient  or 
difficult  to  find  direct  proof.  In  Prop.  XXVII.,  for  exam- 
ple, we  show  that  CD  must  coincide  with  CE  because  their 
noncoincidence  would  entail  a  consequence  that  had  been 
shown  to  be  impossible.  Instead  of  proving  directly  the 
conclusion  we  wish  to  establish,  we  prove  it  indirectly  by 
showing  that  any  other  conclusion  would  lead  to  a  contra- 
diction of  some  known  truth. 

Among  the  exercises  about  to  be  given,  most  are  so  easy 
that  the  principles  on  which  to  base  a  synthetic  proof  at 
once  suggest  themselves.  The  student  should,  however,  in 
every  case,  base  his  synthetic  proof  upon  a  previous  analy- 
sis, guided  by  the  general  directions  given  below.  The 
diagrams  and  references  given  as  aids  should,  as  far  as 
possible,  be  left  as  a  last  resort. 


72  PLANE   GEOMETRY.  — BOOK  I. 

ANALYSIS    OF    THEOREMS. 

1.  Assuming  the  theorem  as  true,  construct  a  diagram, 
accordingly. 

2.  Deduce,  with  the  aid  of  SKch  constructions  as  may  he 
necessary,  any  consequences  that  follow  from  that  assumption, 
a}iplying  such  theorems  already  proved  as  are  ajyplicahle  to  the 
diagram. 

3.  A  consequence  that  contradicts  some  known  truth  shoivs 
the  theorem  to  be  false,  and  ive  can  proceed  to  prove  it  so  by  a 
direct  ptroof. 

4.  On  arriving  at  a.  consequence  known  to  be  true,  take 
this  as  the  basis  of  a  synthetic  proof,  retracing  the  steps  by 
tvhich  this  consequence  was  reached. 

Oiveii,  for  example,  the  following  theorem : 

153.  Theorem.  TJie  bisectors  of  the  angles  of  a  triangle 
meet  in  a  point. 


A 


Given:  In  triangle  ABC,  AD,  BE,  CF,  the  bisectors  of  angles 
A,  B,  C,  respectively ; 

To  Prove :     AD,  BE,  CF,  have  a  common  point. 

Analysis.  Let  O  be  the  point  common  to  the  three 
bisectors  AD,  BE,  CF ;  then  0  must  be  equidistant  from 
AB,  AC,  and  BC  (101).  But  this  condition  is  satisfied  if 
there  is  a  point  common  to  BE  and  CF,  since  that  point 
must  be  equidistant  from  AB  and  BC,  from  BC  and  AC  (101). 
Now,  in  order  that  BE  and  CF  may  meet  in  a  point,  we 
must  have  /.  EBC  +  Z.  FCB  <  a  straight  angle  (114).     But 


GEOMETRICAL   SYNTHESIS  AND  ANALYSIS.     73 

these  angles  are  less  than  a  straight  angle,  since  they  are  the 
halves  of  A  ABC,  ACB,  respectively,  which  are  less  than  a 
straight  angle  (84).     Hence 

Synthesis.     Since  Z  B  -\-  Z  c  <  a  st.  Z,  (84) 

Z  EBC  +  Z  FOB  <  a  St.  Z, 
.'.  BE,  CF,  will  meet  in  a  point  0.  (H^) 

Now,  since  0  is  in  the  bisector  BE  of  Z  ABC, 

0  is  equidistant  from  BA  and  BC ; 
similarly  0  is  equidistant  from  CA  and  BC, 
.'.  o  is  in  the  bisector  AD  of  Z  BAC,     q.e.d.     (101") 
(since  it  is  equidistant  from  AB  and  AC.) 


(101') 


154.  Scholium.  The  point  of  intersection  of  the  bisec- 
tors of  any  two  angles  of  a  triangle  is  equidistant  from  all 
the  sides. 

155.  Theorem.  The  perpendiculars  at  the  mid  points  of 
the  sides  of  a  triayiyle  meet  in  a  poiiit. 


Given:  In  triangle .45(7,  DE,  FG,  HK,  perpendiculars  to  AB, 
AC,  BC,  at  their  respective  mid  points ; 

To  Prove :  DE,  FG,  HK,  have  a  common  point. 

The  analysis  and  synthesis  of  this  theorem  are  so  similar 
to  those  of  the  preceding  theorem  that  they  may  be  fairly 
left  for  the  student  to  supply,  with  the  remark  that  the 
analysis  depends  upon  Prop.  XXIII.  and  its  first  Cor.  much 
as  that  of  the  preceding  theorem  depends  upon  Prop.  XXV. 

/'  OF  THE  \ 

I  l^N/VERsiTYf 


74 


PLANE    GEOMETRY.— BOOK  I. 


156.  Scholium.  The  point  of  intersection  of  any  two  of 
the  perpendiculars  at  the  mid  points  of  the  sides  of  a 
triangle  is  equidistant  from  all  the  vertices. 

157.  Definitions.  In  any  triangle,  the  perpendicular 
from  a  vertex  to  the  opposite  side  is  called  the  altitude 
to  that  side;  the  line  joining  a  vertex  with  the  mid  point 
of  the  opposite  side  is  called  the  mediaii  to  that  side; 
the  sum  of  the  sides  is  called  the  perimeter.  In  an  isosceles 
triangle,  the  equal  sides  may  be  referred  to  as  the  arms, 
and  the  other  side  as  the  base. 


EXERCISES. 
THEOREMS. 

112.  If  the  sides  of  a  right  angle  BAC  are 
produced  through  A^  the  new  angles  thus 
formed  are  all  right  angles.     (30,  50) 

113.  The  bisectors  of  adjacent  supple- 
mentary angles  are  perpendicular  to  each 
other.     (46) 

114.  Conversely,  if  the  bisectors  of  two 
adjacent  angles  are  perpendicular  to  each 
other,  those  angles  are  supplementary.     (46) 

115.  The  bisectors  of  vertical  angles  are  in 
the  same  straight  line.     (49) 

116.  If  a  line  bisect  one  of  two  vertical 
angles,  it  bisects  the  other  also.     (50) 

117.  The  intercepts  bisecting  the  base 
angles  of  an  isosceles  triangle  are  equal.     (63) 

118.  Any  intercept  drawn  perpendicular  to 
the  bisector  of  an  angle  cuts  off  equal  parts 
from  the  sides.     (63) 

119.  The  altitudes  to  the  arms  of  an  isosce- 
les triangle  are  equal.     (63) 


C 

A B 


EXERCISES. 


76 


(66) 


120.  The  bisectors  of  the  base  angles  of  an  isosceles 
triangle  form,  if  they  meet,  an  isosceles  triangle.    (65) 

121.  The  medians  to  the  arms  of  an  isosceles  trian- 
gle are  equal.     {66) 

122.  Enunciate  and  prove  the  converse  of  Exer- 
cise 118.     (66) 

123.  Enunciate  and  prove  the  converse  of  Exer- 
cise 120.     (66) 

124.  Lines  drawn  from  the  extremities  of  the  base 
of  an  isosceles  triangle  to  points  equally  distant  from 
the  vertex  are  equal,  and  divide  the  arms  into  parts 
that  are  mutually  equal.     {66) 

125.  Lines  drawn  from  the  vertex  of  an  isosceles 
triangle  to  points  in  the  base  equally  distant  from  its 
extremities  are  equal.     (66) 

126.  If  equal  distances  from  the  vertices  of  an 
equilateral  triangle  be  laid  off  in  the  same  order,  the 
lines  joining  these  points  form  an  equilateral  triangle. 

127.  If  the  diagonals  of  a  quadrilateral  bisect  each  other,  the  figure 
is  a  parallelogram.     (66) 

128.  If  a  quadrilateral  has  two  adjacent 
sides  equal  and  making  equal  angles  with  the 
other  two  sides,  these  are  also  equal.     (68) 

129.  An  equilateral  triangle  is  equiangular ; 
and  conversely.     (68,  65) 

130.  Prove  that  in  the  diagram  for  Exercise  128,  the  diagonals 
divide  the  figure  into  pairs  of  equal  triangles.    (69) 

131.  The  perpendiculars  to  a  diagonal  of  a 
parallelogram  from  the  opposite  vertices  are 
equal.     (73) 

132.  The  diagonals  of  a  square  or  rhom- 
bus bisect  each  other  at  right  angles,  and  bisect 
the  angles  whose  vertices  they  join.     (74) 

133.  If  from  any  point  P  in  A  ABC,  PB, 
PC,  be  drawn,  then  Z  P>  Z^l,  but 

PB  +  PC<AB  +  AC.     (83,  88) 


76 


PLANE   GEOMETRY.— BOOK   L 


134.  If  in  a  quadrilateral  the  greatest  side  is 
opposite  the  least,  each  angle  including  the  least 
side  is  greater  than  the  opposite  angle.     (86,  88) 

135.  The  sum  of  the  lines  drawn  from  any  point 
in  a  triangle  to  the  vertices  is  less  than  the  sum, 
but  greater  than  the  semisum,  of  the  sides.     (88) 

136.  In  a  triangle  ABC^  if  AB  is  not  less  than 
AC,  any  intercept  AD  between  A  and  J5C  is  less 
than  AB.     (87) 

137.  In  a  triangle  ABC,  a  median  AD  is  drawn 
to  BC.  Show  that  the  angle  ADB  is  obtuse,  right, 
or  acute,  according  as  AB  >,  =,  or  <  AC.    (91) 

138.  The  median  to  any  side  of  a  triangle  is 
less  than  the  semisum  of  the  other  two  sides, 
but  greater  than  the  semidifference  of  their  sum 
and  the  third  side.     (88) 

139.  A  line  drawn  from  any  point  in  the  bisec- 
tor of  an  angle,  parallel  to  one  side,  meets  the  other 
in  a  point  equidistant  from  the  vertex  and  the 
point.     (110) 

140.  If  AB.,  an  arm  of  the  isosceles  triangle 
ABC,  be  produced  so  that  AD  =  AB,  then  will 
the  line  joining  Z>C  be  perpendicular  to  BC    (120) 

141.  If  BD  is  the  altitude  upon  AC,  one  of  the 
arms  of  the  isosceles  triangle  ABC,  then  the  angle 
DBC  is  equal  to  one  half  the  vertical  angle.    (120) 

142.  The  sum  of  the  medians  to  the  sides  of  a 
triangle  is  less  than  the  sum,  but  greater  than  the 
semisum,  of  the  sides.     (Exercise  138) 

143.  If  AP  is  the  altitude,  and  AM  the  bisec- 
tor, from  A  to  BC,  then 

ZPA3f=l(ZC~ZB).      (120)  B'         MFC 

144.  The  median  to  the  hypotenuse  of  a  right  triangle  is  equal  to 
one  half  the  hypotenuse.     (120) 


EXERCISES. 


77 


145.  If  the  bisectors  of  two  angles  of  an  equilateral  triangle  meet, 
and,  from  the  point  of  meeting,  lines  be  drawn  parallel  to  any  two. 
sides,  these  lines  will  trisect  the  third  side.     (120) 

146.  The  bisector  of  an  exterior  angle  at 
the  vertex  of  an  isosceles  triangle  is  parallel 
to  the  base.     (122) 

147.  If,  in  a  triangle  ABC,  the  bisectors 
of  an  interior  angle  at  B  and  of  an  exterior 
angle  at  C  meet  in  D,  then  ZD  =  ^ZA. 

148.  The  angle  formed  by  the  bisectors 
of  any  two  consecutive  angles  of  a  quadri- 
lateral is  equal  to  the  sum  of  the  other  two 
angles.     (125) 

149.  The  sum  of  the  distances  of  any  point 
in  the  base  of  an  isosceles  triangle  from  the 
arms,  is  constant ;  i.e.,  is  always  equal  to  the 
altitude  upon  an  arm.     (143) 

150.  The  sum  of  the  distances  of  any  point 
within  an  equilateral  triangle  from  the  sides 
is  constant;  i.e.,  is  equal  to  an  altitude.  (Exer- 
cise 148) 

151.  The  lines  joining  the  mid  points  of 
the  sides  of  a  triangle  divide  it  into  four 
equal  triangles.     (152) 

152.  The  three  altitudes  of  a  triangle  have 
a  common  point.     (141,  155) 

153.  The  lines  joining  the  mid  points  of 
adjacent  sides  of  any  quadrilateral,  form  a 
parallelogram  the  sum  of  whose  sides  is 
equal  to  that  of  the  diagonals  of  the  quad- 
rilateral.    (147) 

154.  The  three  medians  of  a  triangle  have 
a  common  point  which  cuts  off  one  third  of 
each  median.     (147) 

155.  If  the  exterior  angles  of  a  triangle  are  bisected,  the  three 
exterior  triangles  formed  on  the  sides  of  the  original  triangle  are  equi- 
angular. 

156.  The  vertices  of  all  right  triangles  having  a  common  base  as 
hypotenuse,  lie  in  the  same  circumference.     (Exercise  144) 


Book  II. 
the  circle.   loci.   problems. 


»«<o 


ELEMENTARY  PROPERTIES. 

158.  A  circle  is  a  plane  figure  bounded  by  a  curved 
line  such  that  a  certain  point  within,  called  the  center,  is 
equidistant  from  every  point  of  the  curve. 

159.  The  circumference   of   a  circle  is 
the  curve  that  bounds  it ;  as  ADEB  C. 

160.  An  arc  is  any  part  of  a  circum- 
ference; as  AD  or  ACE. 

161.  A  radius  is  any  straight  line  drawn  from  center  to 
circumference;  as  OA. 

162.  CoR.  All  radii  of  the  same  circle  are  equal  (158). 
Also,  a  poi7it  is  within,  on,  or  without  a  circumference  accord- 
ing as  its  distance  from  the  center  is  less  than,  equal  to,  or 
greater  than  a  radius. 

163.  A  chord  is  a  straight  line  joining  the  extremities  of 
an  arc ;  as  DE.     The  chord  is  said  to  subtend  its  arc. 

164.  A  diameter  is  a  chord  that  passes  through  the 
center ;  as  AB. 

165.  CoR.  All  diameters  of  the  same  circle  are  equal. 
For  each  is  double  the  radius  of  that  circle. 

78 


ELEMENTARY  PROPERTIES.  79 

Proposition  I.     Theorem. 

166.  The  straight  line  terminating  in  any  two 
points  in  a  circumference  lies  wholly  within  the 
circle. 


Given :  A  straight  line,  AB,  terminating  in  two  points  in  the 
circumference  whose  center  is  0  ; 

To  Prove :  AB  lies  within  that  circumference. 

Take  any  point  P  between  A  and  B,  and  join  OA,  OB,  OP. 
Since  OA  =  OB,  (162) 

Zb  =  Z.A.  (68) 

But  /.A<  ext.  Z  OPB,  •  (83) 

.-.  Zb<Zopb, 

.'.    OP<OB,  (87) 

i.e.,  P,  which  is  any  point  between  A  and  B,  lies  within  the 
circle.  q.e.d. 

167.  Cor.  A  straight  line  can  meet  a  circumfereyice  in  not 
more  than  two  points. 

Exercise  157.  Show  that  a  circle  cannot  have  more  than  one 
center.    State  the  axioms  upon  which  your  proof  depends. 

158.  A  straight  line  will  cut  a  circle,  or  lie  entirely  without  it, 
according  as  its  distance  from  the  center  is  less  than,  or  greater  than, 
the  radius  of  the  circle. 

159.  If  a  straight  line  could  meet  a  circumference  in  three  points, 
how  many  equal  straight  lines  could  be  drawn  from  the  center  to  that 
line? 


80  PLANE   GEOMETRY.  — BOOK  11. 

Proposition  II.     Theorem. 
168.  Circles  having  equal  radii  are  equal. 


Given:  Two  circles,  ABD,  a'b'b',  having  radius  OA  equal  to 
radius  o'A' ; 

To  Prove:     Circle  ABB  is  equal  to  circle  a'b'b'. 

Place  Oabb  upon  O  a'b'b',  so  that  OA  ^  o'aK 

Then,  not  only  will  A  coincide  with  A',  but  also  circum- 
ference ABB  with  circumference  A'b'b',  since  all  points  of 
both  circumferences  are  equally  distant  from  the  coinciding 
centers  0  and  O'  (Hyp.  and  158).     Hence, 

since  circumf.  ABB  =^  circumf.  A'b'b', 

Q  ABB  =  0  A'b'b'.  q.e.d.     (14) 

169.  Cor.  A  diameter  bisects  the  circle  and  its  circum- 
ference. 

For  producing  AO,  A'o',  to  form  the  diameters  AC,  A'c' ; 
since  the  part  .4i?C  may  be  made  to  coincide,  as  above,  either 
with  A'b'c'  or  a'b'c',  these  parts  are  equal,  as  are  also 
their  bounding  arcs.  (Ax.  1) 

170.  Defixition.  a  half  circle  is  called  a  semicircle; 
a  half  circumference,  a  semicircumference. 

Exercise  160.  How  many  points  are  necessary  to  determine  the 
magnitude  and  position  of  a  circle  in  a  given  plane  ? 

161.    Can  two  circles  have  a  common  center  without  coinciding? 


CHORDS.  81 

CHORDS. 

Proposition  III.     Theorem. 

171.   If  a  chord  is  perpendicular  to  another  chord 
at  its  mid  point,  the  fij^st  chord  is  a  diameter. 


Given :  Chord  AB  perpendicular  to  chord  CD  at  its  mid  point  E  ; 
To  Prove :  AB  is  Si  diameter. 

Since  AB  is  _L  to  CD  at  its  mid  point,  (Hyp.) 

AB  contains   all  the   points   that   are   equidistant    from 

C  and  D ;  (96) 

.'.  the  center  of  Q  ACD  lies  in  AB,  (158) 

.-.  AB  is  a  diam.  of  Q)  ACD.      q.e.d.     (164) 

172.  Cor.  A  radius  perpendicular  to  a  chord  bisects  that 
chord.  (97) 

Exercise  162.  In  the  diagram  for  Prop.  III.,  if  AC,  AD,  be  joined, 
then^C=  AD. 

163.  The  line  drawn  through  the  mid  points  of  parallel  chords  in  a 
circle,  passes  through  the  center. 

164.  If  an  isosceles  triangle  be  constructed  on  any  chord  of  a  circle, 
its  vertex  will  be  in  a  diameter  or  a  diameter  produced. 

165.  If  two  chords  of  a  circle  bisect  each  other,  both  are  diameters. 

166.  Enunciate  the  converse  of  Prop.  III.  Is  that  converse  always 
true  ? 

167.  If  from  any  point  within  a  circle  two  equal  straight  lines  be 
drawn  to  the  circumference,  the  bisector  of  the  angle  they  form  will 
pass  through  the  center. 

Geom.  — 6 


82  PLANE   GEOMETRY.  — BOOK  11. 

Proposition  IV.     Problem. 
173.    To  find  the  center  of  a  given  circle. 


c 


Given :  A  circle  A  CB  ; 

Required :  To  find  its  center. 

Join  any  two  points  A,  B,  in  the  circnmf. 

Draw  CD  J_  to  ^5  at  its  mid  point.  (95) 

Bisect  CD  in  0  (80).     0  is  the  required  center. 

For  chord  CD  is  ±  to  AB  at  its  mid  point,    (Const.) 

•    .-.  the  center  of  Q)  ACB  lies  in  CD ;  (I'i'l) 

and  since  0(7=  OD,  (Const.) 

0  is  the  center  of  O  ACB,  q.e.f. 

(it  being  the  mid  point  of  the  diam.  CD.) 

Scholium.  In  order  to  avoid  repetition,  it  will  usually 
be  assumed,  henceforth,  that  the  center  0  of  any  circle, 
if  not  given,  has  been  found  by  this  construction. 

Exercise  168.  Show  how  to  find  the  center  of  a  circle,  when  an 
arc  only  of  its  circumference  is  given. 

169.  Apply  Exercise  167  so  as  to  find  the  center  of  a  circle  without 
the  bisection  of  any  chord . 

170.  Apply  Exercise  168  to  effect  the  same  purpose. 

171.  Find  the  direction  in  which  lies  the  longest,  and  also  the 
shortest,  distance  from  a  point  within  a  circle  to  its  circumference. 

172.  Find  the  same  for  a  point  without  the  circle. 


CHORDS. 


83 


Proposition  V.     Theorem. 

174.   In  the  same  circle,  or  in  equal  circles,  equal 
arcs  are  subtended  by  equal  chords;  and  conversely. 


1°.    Given:  In  equal  circles  ADB,  a'd'b\  AB,  A'b',  chords 
of  equal  arcs  ACB,  a'c'b' ; 

To  Prove :     Chord  AB  is  equal  to  chord  A'b'. 

0,  O',  being  the  centers,  place  O  ADB  upon  Qa'd'b', 
so  that  0=^0'  and  A=^A'.     Then 

since  arc  ACB  =  arc  A'c'b',  (Hyp.) 

B  will  coincide  with  B', 
.',  chd.  AB  will  coincide  with,  and  equal,  chd.  A'b'.  q.e.d.  (14) 

2°.    Given:  In  equal  circles  ADB,  a'd'b',  equal  chords  AB, 
A^B'; 

To  Prove :     Arc  ACB  is  equal  to  arc  A'c'b'. 
Join  OA,  OB,  O'A',  O'b'.     Then 
since  OA  =  0'A',  OB  =  o'b',  and  AB  =  a'b',    (Hyp.) 
AOAB  =A0'a'b',  (69) 

.-.  O  ADB  can  be  placed  upon  Q  a'd'b',  so  that 

A0AB=^A0'A'B';  (61) 

then  arc  ACB  4^  arc  A'c'b', 
.:  SiTC  ACB  =  SiTG  A'c'b'.         q.e.d.     (14) 

Scholium.   If  the  arcs  are  in  the  same  circle,  the  proof 
is  similar  to  that  above. 


84  PLANE   GEOMETRY.  — BOOK  11. 

Proposition  VI.     Theorem. 

175.   A  radius  bisectirig  a  chord  bisects  also   its 
subtended  arc. 

D 


Given :  A  radius  OC  bisecting  chord  AB ; 

To  Prove :  Arc  BCA  is  bisected  in  c. 

Join  CA  and  CB.     Then 
since  0(7  is  ±  to  .4^  at  its  mid  point,         (Hyp.) 
chd.  CA  =  chd.  CB,  (96) 

.-.  arc  (7J  =  arc  C£.  q.e.d.     (174) 

176.  Cor.  1.  A  radius  bisecting  an  arc  bisects  its  chord 
at  right  angles.  .  (75) 

For  the  extremities  of  the  radius  are  equally  distant  from 
those  of  the  chord. 

177.  Cor.  2.  A  radius  bisecting  a  chord  or  its  subtended 
arc  is  perpendicidar  to  the  chord.  (75  and  176) 

Exercise  173.  In  the  diagram  for  Prop.  V.,  if  the  bisector  of 
angle  ^OB  be  drawn,  it  will  bisect  arc  A  CB  and  also  arc  ADB. 

174.  In  the  same  diagram,  if  the  points  where  the  bisector  cuts  the 
circumference  be  joined  with  A  and  B,  the  chords  drawn  from  each 
point  will  be  equal. 

175.  In  the  diagram  for  Prop.  VI.,  if  diameters  from  A  and  B  meet 
the  circumference  in  E  and  F,  the  line  joining  EF  will  be  parallel 
to  AB. 

176.  Two  intersecting  circles  cannot  have  the  same  center. 


CHORDS.  85 

Proposition  VII.     Problem. 

178.  In  a  given  circle,  to  place  a  chord  equal  to  a 
given  straight  line  not  greater  than  a  diameter. 


M 


J! 


Given :  A  straight  line  MN  not  greater  than  the  diameter  of  a 
circle  adb  ; 

Required:  To  place  m  ADB  a  chord  equal  to  MN. 

From  A,  any  desired  point  in  the  circumference,  as  center, 
with  radius  equal  to  MN,  describe  an  arc  CB,  cutting 
the  circumference  in  B. 

Join  AB.     AB  i^  the  required  chord. 
Since  AB  is  a  radius  of  arc  BC,  and  AB  =  MN,     (Const.) 
a  chd.  AB  equal  to  MN  has  been  placed  in  O  ADB.     q.e.f. 

179.  Scholium.  By  this  construction,  any  given  arc  may 
be  laid  off  on  a  circumference  having  a  radius  equal  to  that 
of  the  arc. 

180.  Definition.    An  arc  is  called  a  major  or  a  minor  arc  | 
according  as  it  is  greater  or  less  than  a  semicircumference. 

Exercise  177.  In  the  diagram  for  Prop.  VII.,  if  arc  EC  be  pro- 
duced to  meet  the  circumference  in  2),  and  BD  be  drawn,  the  diameter 
drawn  from  A  will  bisect  BD. 

178.  In  the  same  diagram,  place  a  chord  that  is  equal  and  parallel 
to  AB. 

179.  In  the  same  diagram,  place  a  chord  that  is  equal  and  perpen- 
dicular to  AB. 


86  PLANE   GEOMETRY.  — BOOK  11. 

Proposition  VIII.     Theorem. 

181.  In  the  same  circle,  or  in  equal  circles,  the 
greater  of  two  minor  arcs  is  subtended  by  a  greater 
chord;  and  conversely- 


1°.  Oiven:  In  circle  ADB,  arc  AFB  greater  than  arc  CED ; 
To  Prove:  Chord  ^^  is  greater  than  chord  CB. 

From  A  draw  chd.  AF  =  CB  (178),  and  join  0,  the  center, 
with  A,  B,  F. 

Since  cM.  ^-F  =  chd.  CD,  (Const.) 

arc  ^7^  =  arc  CEB  (174")  and  <  arc  AFB,      (Hyp.) 

,-.  F  falls  between  A  and  B,  and  OF  between  OA  and  OB, 

.:  Zaob>  Zaof.  (Ax.  8) 

Now  OA,  OB  —  OA,  OF,  respectively,  (162) 

.'.  AB  >AF  or  CB.  Q.E.D.     (90) 

2°.    Given:  In  circle  ADB,  chord  AB  greater  than  chord  CD ; 
To  Prove :  Arc  AFB  is  greater  than  arc  CED. 
With  the  same  construction  as  in  1°, 
since  OA,  OB  =  oA,  OF,  respectively,  (162) 

but  AB>AF,  (Hyp.) 

Zaob>Zaof,  (90) 

.-.  F  falls  between  A  and  B, 
,\  arc  AFB  >  arc  AF  or  arc  CD.     q.e.d.     (Ax.  8) 


CHORDS.  87 

Scholium.  If  the  arcs  are  in  equal  circles,  the  proof  is 
similar  to  that  above ;  and,  in  general,  the  demonstration 
of  a  theorem  concerning  arcs,  etc.,  in  equal  circles  would  be 
similar  to  that  for  arcs  in  the  same  circle,  and  vice  versa. 


Propositiox  IX.     Theorem. 

18^.  In  the  same  circle,  or  in  equal  circles,  equal 
chords  are  equally  distant  froin  the  center;  and 
conversely. 


1°.    Given :  In  circle  ADB,  chord  AB  equal  to  chord  CD  ; 
To  Prove :  AB  and  CD  are  equally  distant  from  0,  the  center. 

Draw  OP  J_  to  ^5,  OR  1.  to  CD,  and  join  OA,  OC. 

Since  OP  is  ±  to  AB,  and  OR  is  ±  to  CD,   (Const.) 

AP  =  1-  AB,  and  CR  =  ^  CD,  (172) 

.-.  AP  =  CR  (Ax.  7),  and  OA  =  OC,  (162) 

.-.  rt.  AOAP  =  rt.  A  OCR,  (72) 

.-.    OP  =  OR.  Q.E.D.       (70) 

2°.    Given :  In  circle  ADB,  chords  AB,  CD,  equally  distant  from 
9,  the  center ; 

To  Prove :  AB  =  CD. 

With  the  same  construction  as  in  1°, 

since  OP  =  OR  (Hyp.),  and  OA  =  00,  (162) 

rt.  A  OAP  =  rt.  A  OCR,  (72) 

.-.  AP=  CR,  (70) 

,-.  2ap  or  AB  =  2CR  OT  CD.    Q.E.D.  (Ax.  6) 


88  PLANE   GEOMETRY.  —  BOOK  II. 

Proposition  X.     Theorem. 

183.  In  the  same  circle,  or  in  equal  circles,  a  greater 
chord  is  nearer  the  center;  and  conversely. 


1°.    Given :  In  circle  ADB,  op  perpendicular  to   chord   AB, 
OQ  perpendicular  to  chord  CD,  and  AB  greater  than  CB  ; 
To  Prove :         OP  is  less  than  OQ. 

Since  AB  >  CD  (Hyp.),  arc  AEB  >  arc  CD,  (181") 

On  arc  AEB  Islj  off  arc  ^^  =  arc  CD.  (1'<'8) 

Join  AE,  and  draw  OR  _L  to  AE  ;  (92) 

then  AE=  CD  (174'),  and  0R=  OQ.  (182') 

Since  AB  lies  between  0  and  AE,  (Const.) 

OR  must  cut  AB  in  some  point  S, 

.'.    OS  <  OR.  (Ax.  8) 

But  OP  <  OS,  '   (93) 

.-.  OP  <0R  or  OQ.  Q.E.D.       (a.f.) 

2°.    Given:  In  circle  ADB,   OP   perpendicular  to  chord  AB, 
OQ  perpendicular  to  chord  CD,  and  OP  less  than  OQ; 
To  Prove:  Chord  AB  is  greater  than  chord  CD. 

Chord  AB  cannot  be  equal  to  chord  CD,  for  then  OP  would 
be  equal  to  OQ  (182')  ;  nor  can  AB  be  less  than  CD,  for  then 
OP  would  be  greater  than  OQ  (1°).  Hence,  since  AB  can  be 
neither  equal  to  nor  less  than  CD,  it  must  be  greater  than  CD. 

184.  Cor.  A  diameter  is  greater  than  any  chord  not  pass- 
ing through  the  center. 


CHORDS.  89 

Proposition  XI.     Theorem. 

185.  Through  any  three  given  points  not  in  the 
same  straight  line  one  circumference  can  be  described, 
arid  only  one. 


Given :  Three  points  A,  B,  C,  not  in  the  same  straight  line; 
To  Prove:  One  circumference  can  be  described  through  A,  B, 
and  C.  , 

Join  AB,  BC,  and  at  D,  E,  the  mid  points  of  AB,  BC,  respec- 
tively, draw  DO  ±  to  AB,  and  EO  ±  to  BC*  (80) 

•••  DO,EO,  are  ±  to  AB,BC,  resp.,  at  their  mid  points,  (Const.) 

DO,  EO,  will  meet  at  a  point  0  equidistant  from  A,  B, 

and  a;  (156) 

.-.  the  circumference  described  from  0  as  center,  with  radius 
OA,  will  pass  through  A,  B,  and  C.  q.e.d.      (162) 

Moreover,  there  can  be  but  one  such  circumference.  For 
the  center  of  any  circumference  passing  through  A,  B,  and 
C,  must  lie  both  in  DO  and  in  EO  (97),  which  lines  can  have 
but  one  common  point. 

186.  CoR.  Two  circumferences  can  intersect  in  not  more 
than  two  points. 

For  if  they  could  intersect  in  three  points,  there  would 
be  two  different  circumferences  passing  through  the  same 
three  points,  which  is  impossible. 

187.  Scholium.  One  circumference,  and  only  one,  can  be 
described  through  the  vertices  of  a  given  triangle. 


*  The  construction  by  wliich  we  find  the  mid  point  of  a  line,  evidently 
also  gives  the  perpendicular  at  that  point. 


90  PLANE    GEOMETRY.  — BOOK  II. 

TANGENTS   AND    SECANTS. 

188.  A  tangent  is  a  straight  line  that  touches  a  circum- 
ference in  one  point  only;   as  ABC.     The 
point  B  in  which  the  tangent  touches  the     ^i 
circle  is  called  the  point  of  contact. 

189.  A  secant  is  a  straight  line  that  cuts 
a  circumference  in  two  points ;  as  DE. 

Proposition  XII.     Theorem. 

190.  A  straight  line  perpendicular  to  a  radius  at 
its  extremity  is  a  tangent  to  the  circle. 


Given:  AB  perpendicular  to  0(7,  a  radius  of  circle  CEF,  at  its 
extremity  C ; 

To  Prove :      AB  is  a  tangent  to  circle  CEF. 

Join  O  with  D,  any  point  in  AB  except  C.     Then 

since  oc  is  ±  to  AB,  (Hyp.) 

OD  is  oblique  to  AB,  (51) 

.-.  0D>  oc,  (93) 

.*.  D  lies  without  the  circle  CEF,  (162) 

.-.  AB  is  tangent  to  CEF  at  C,    q.e.d.    (188) 

(since  C  is  the  only  point  in  AB  not  without  CEF.) 

191.  Cor.  1.  A  tangent  is  perpendicular  to  the  radius 
drawn  to  the  point  of  contact. 

For  OC  must  be  less  than  any  other  line  drawn  from 
0  to  AB  (162). 


TANGENTS  AND  SECANTS. 


91 


192.  Cor.  2.    A  perpendicular  to  a  tangent  at  the  point  of 
contact  passes  through  the  center  of  the  circle. 

For  otherwise  there  could  be  two  perpendiculars  to  the 
tangent  at  that  point  (191). 

193.  CoR.  3.    A  perpendicular  from  the  center  to  a  tangent 
meets  it  at  the  point  of  contact  (192). 

194.  CoR.  4.   At  a  given  point  of  contact  there  can  he  hut 
one  tangent. 


Proposition  XII  I.     Theorem. 

195.   Parallels  intercept  equal  arcs  on  a  circum- 
ference. 


Given:  Two  parallels  AB,   CD,  cutting  or  touching  the  cir- 
cumference whose  center  is  0  ; 

To  Prove:  These  parallels  intercept  equal  arcs. 

1°.   Let  AB,  CD,  be  secants. 

Through  0,  the  center,  draw  the  diameter  MN  _L  to  AB. 

Since  AB  is  II  to  CD  (Hyp.),  and  MN  is  _L  to  AB,   (Const.) 

MN  is  ±  to  CD,  (107) 

.-.  arc  MA  =  arc  MB,  and  arc  MC  =  arc  MD,       (175) 

.-.  arc  ^C  =  arc  i5Z).       q.e.d.        (Ax.  3) 

2°.   Let  AB  be  tangent  at  M,  and  CD  a  secant. 

Draw  OM.     Then  OM  is  ±  to  AB  (191) 

and  also  to  its  parallel  CD  ;  (107) 

.-.  arc  i^/C  =  arc  J/JD.  q.e.d.  (175) 


92  PLANE   GEOMETRY.  — BOOK  IL 

3°.   Let   AB,  CD,  be   tangents  at  M,  N, 

respectively.  ^e/^^^\^^^f 

Draw  a  secant  EF  II  to  AB  ;  it  will  also  "/ V' 

be  II  to  en.  (108)  (        ^  1 

Since  arc  me  =  ^yc  MF,  and  arc  NEz=       \  / 

arc  NF,  (2°)  c^^N^D 

arc  3fEN  =  arc  MFN.  q.e.d.    (Ax.  2) 

196.  Definition.  The  line  joining  the  centers  of  two 
circles  is  their  Zme  of  centers;  the  distance  between  their 
centers  is  their  central  distance. 

197.  Two  circles  are  tangent  to  each  other  when  both 
are  tangent  to  the  same  straight  line  at  the  same  point. 

198.  A  circle  tangent  to  another  is  tangent  to  it  internally 
or  externally  according  as  it  lies  within  or  without  the  other 
circle. 


Proposition  XIV.     Theorem. 

199.   If  two  circles  are  tangent  to  each  other,  their 
line  of  centers  passes  through  the  point  of  contact. 


Given:  Circles  with  centers  0,  O',  tangent  at  P  in  ^5; 
To  Prove :       P  lies  in  the  line  joining  oo'. 

Through  P  draw  a  perpendicular  to  AB.  (95) 

Since  ^^  is  tangent  to  both  circles  at  P,      (Hyp.) 

the  J_  to  AB  at  P  passes  through  O  and  O',     (192) 

i.e.,  P  is  in  the  line  of  centers  00'.  q.e.d. 


EXERCISES,  93 

EXERCISES. 
QUESTIONS. 

180.  Can  you  find  the  center  of  a  circle  without  bisecting  any 
straight  line  ? 

181.  If  a  chord  of  one  circle  is  equal  to  a  chord  of  another,  are 
these  circles  necessarily  equal  ?  If  not,  what  other  condition  must  be 
fulfilled  that  the  circles  may  be  equal  ? 

182.  In  the  diagram  for  Prop.  VII.,  if  iHfiVwere  equal  to  a  diame- 
ter, through  what  point  would  AB  necessarily  pass  ? 

183.  In  equal  circles,  which  is  subtended  by  the  greater  chord,  the 
greater  or  the  less  of  two  major  arcs  ? 

184.  Can  a  tangent  be  drawn  to  a  circle  from  a  point  within  it  ? 

185.  How  may  a  tangent  be  drawn  at  a  given  point  in  a  circum- 
ference when  the  center  is  not  known  ? 

186.  How  many  points  determine  a  circumference  ? 

187.  Is  there  any  limit  to  the  number  of  circles  that  may  be 
described  through  two  given  points  ? 

188.  In  what  case  is  it  impossible  to  pass  a  circle  through  three 
given  points  ? 

189.  Can  a  circumference  always  be  described  through  the  angular 
points  of  a  rectangle  ? 

190.  How  many  circles  of  equal  radii  may  touch  a  given  straight 
line  at  a  given  point  ? 

191.  How  many  circles  of  any  radii  may  touch  a  given  straight  line 
at  a  given  point  ? 

THEOREMS. 

192.  A  circle  is  wholly  without  or  within  another  circle,  according 
as  their  central  distance  is  greater  than  the  sum,  or  less  than  the  dif- 
ference, of  their  radii.     (162) 

193.  A  circle  is  tangent  to  another  externally  or  internally,  accord- 
ing as  their  central  distance  is  equal  to  the  sum  or  the  difference  of 
their  radii.     (199) 

194.  Two  circles  intersect  if  their  central  distance  is  less  than  the 
sum,  and  greater  than  the  difference,  of  their  radii.     (88,  89)^ 


94  PLANE    GEOMETRY.  — BOOK   IL 

195.  If  two  circles  intersect,  their  line  of  centers  is  perpendicular 
to  their  common  chord  at  its  mid  point.     (172) 

196.  The  center  is  the  only  point  from  which  more  than  two  equal 
lines  can  be  drawn  to  the  circumference.     (97,  171) 

197.  If  two  equal  chords  intersect,  their  segments  are  severally 
equal.     (174,  69) 

198.  If  through  any  point  in  a  radius  two  chords  be  drawn  making 
equal  oblique  angles  with  it,  these  chords  are  equal.     (182") 

,199.  The  chord  drawn  through  any  point  of  a  radius  perpendicular 
to  it  is  the  least  chord  that  can  be  drawn  through  that  point.     (183) 

200.  The  tangents  drawn  to  a  circle  from  any  point  without  it  are 
equal,  and  make  equal  angles  with  the  line  joining  the  point  with  the 
center.     (192) 

201.  The  sums  of  the  opposite  sides  of  a  quadrilateral  described 
about  a  circle  are  equal.     (Exercise  200) 

202.  Tangents  AD,  BC,  at  the  extremi- 
ties of  a  diameter  AB,  meet  another  tangent 
CD  in  C  and  D;  join  00,  OD.  (1)  COD 
is  a  right  angle  ;  {2)  CD  =  AD  +  BC. 
(Exercise  200) 

203.  If  through  the  points  of  intersection 
of  two  circumferences,  parallels  be  drawn  to  meet  the  circumferences, 
these  parallels  will  be  equal.     (195,  174) 

204.  If  two  tangents  drawn  to  a  circle  from  the  same  point,  inter- 
cept between  them  a  third  tangent  touching  any  point  of  the  inter- 
cepted arc,  the  perimeter  of  the  triangle  formed  by  the  three  tangents 
is  constant.     (Exercise  200) 

205.  If  a  circle  is  inscribed  in  a  triangle,  the  distances  of  the  vertex 
of  any  angle  to  the  points  of  contact  of  its  sides  are  equal  to  the  semi- 
perimeter  of  the  triangle  less  the  side  opposite  the  angle.    (Exercise  200) 

206.  If  a  circle  is  inscribed  in  a  trapezoid  that  has  equal  angles  at 
its  base,  each  nonparallel  side  is  equal  to  half  the  sum  of  the  parallel 
sides.     (Exercise  200) 

207.  If  two  circles  are  each  tangent  to  a  pair  of  parallel  lines  and 
also  to  a  transverse  intercept  between  the  parallels,  the  intercept  is 
equal  to  the  central  distance  of  the  circles.     (Exercise  200) 


CONSTRUCTIONS. 


95 


CONSTRUCTIONS. 

Certain  problems  of  construction  have  already  been  given 
as  occasion  arose  for  their  use.  The  form  in  which  they 
were  presented  was  not,  however,  always  the  simplest  pos- 
sible, but  such  as  the  means  of  demonstration  at  that  stage 
admitted.  Those  already  given  are,  accordingly,  presented 
again  in  a  simplified  form,  with  the  directions  for  their  con- 
struction worded  as  briefly  as  possible.  The  demonstra- 
tions the  student  should  complete  for  himself  with  the  aid 
of  the  diagrams  and  references. 

200.    To    bisect   a   given   straight  c .. 


^  or  arc,  AB. 

From  A  and  B,  with  any  radius 
greater  than  half  the  distance  AB, 
describe  arcs  intersecting  in  C  and  D. 

The  line  CD  bisects  the  line  AB  va. 
X,  and  the  arc  AB  in  y  (74,  175). 


i)- 


201.  To  bisect  a  giveyi  angle  BAG. 
From  A,  with  any  radius,  describe 

an  arc  cutting  AB,  AC,  in  D,  E,  re- 
spectively. From  D  and  E,  with 
the  same  radius,  describe  arcs  inter- 
secting in  X. 

The  line  AX  is   the   bisector   of 
Abac  (101"). 

202.  To  draw  a  perpendicular  to  a  given  line  AB  through 

a  point  C  hi  or  without  AB. 

...  X  .. 
From  C,  with  any  suitable  radius, 

describe  an  arc  cutting  AB  inD  and 

E.    From  D  and  E,  with  equal  radii  / 

of    suitable    length,    describe    arcs       ^ — 

intersecting  in  X. 

The   line  CX   is  perpendicular  to  AB  (74). 


96  PLANE   GEOMETRY.  — BOOK  IL 

It  will  be  seen  that  the  simplification  in  these  cases  arises 
from  the  use  of  arcs  of  any  convenient  radius  instead  of 


whole  circumferences  of  prescribed  radius,  and  in  the  omis- 
sion of  construction  lines  needed  only  for  demonstration. 


Proposition  XV.     Problem. 

203.  At  a  point  C,  in  a  given  straight  line  AB,  to 
construct  an  angle  equal  to  a  given  angle  0. 


F  JJ 

Given:  A  point  C  in  a  straight  line  AB,  and  an  angle  0 ; 
Required :  To  make  at  C  an  angle  equal  to  angle  0. 

From  0  as  center,  with  any  radius,  describe  an  arc  cutting 

the  sides  of  Z  0  in  D  and  E. 
From  C  as  center,  with  the  same  radius,  describe  an  arc 
FG,  and  on  FG  set  off  an  arc  FX  equal  to  arc  DE.  (179) 
Join  ex.     FCX  is  the  required  angle. 
Join  FX.     Then 
since  FX  and  DE  are  equal  arcs  of  equal  circles,   (Const.) 
chord  FX  =  chord  DE  ;  (174') 

also  CF  and  CX  =  OD  and  OE,  resp.  (Const.) 

.-.  AFCX  =  Adoe.     .-.  Za  =  Zo.  Q.E.F. 


CONSTRUCTIONS.  97 

Proposition  XVI.     Problem. 

204.  Through  a  given  point  to  draw  a  parallel  to  a 
^iven  line. 


Given  :  A  straight  line  AB  and  a  point  C ; 

Required:  To  draw  tbrougli  C  a  parallel  to  AB. 

Through  C  draw  ECD,  making  any  Z  D  with  AB. 
At  the  point  C  in  ED  make  Z  ECF  =  /.B.  (203) 

F^QF  is  the  required  parallel. 
For  since  Ac  =  Ab,  (Const.) 

F^CF  is   II  to  AB.  Q.E.F.  (112') 

Exercise  208.  The  common  chord  of  two  intersecting  circles  may 
he  a  diameter  of  one,  hut  not  of  hoth. 

209.  The  chords  joining  the  extremities  of  equal  parallel  chords 
of  the  same  circle  form  with  them  a  rectangle. 

210.  In  the  diagram  for  Prop.  IX.,  if  BA,  DC,  be  produced  to  meet 
in  Q,  the  line  joining  OQ  will  bisect  /.BQD. 

211.  Hence  show  that,  in  the  diagram  for  Prop.  VIII.,  if  AC,  DF, 
be  joined,  AC  will  be  parallel  to  DF. 

212.  In  the  diagram  for  Prop.  XI.,  show  that  Z  0  is  the  supplement 
of  Z  J5. 

213.  In  the  same  diagram,  if  OD,  OE,  be  produced  to  meet  the 
circumference  in  i^  and  G,  then  will  arc  FBG  =  ^(arc  AB  +  arc  BC). 

214.  In  the  same  diagram,  if  UK  be  tangent  to  the  circle  at  B, 
then  will  Z  HBD  +  Z  KBE  be  equal  to  Z  O. 

215.  Tangents  at  the  extremities  of  a  diameter  are  parallel,  and 
conversely.  If  two  radii  are  at  right  angles  to  each  other,  the  tangents 
at  their  extremities  are  perpendicular  to  each  other. 

Geom.  —  7 


98  PLANE   GEOMETRY.  — BOOK  11. 

Proposition  XVII.     Problem. 

205.  To  construct  a  triangle  having  its  sides  respec- 
tively equal  to  three  given  straight  lines,  each  less 
than  the  sum  of  the  other  two. 


Given :  Three  straight  lines,  A,  B,  c,  each  less  than  the  sum  of 
the  other  two ; 

Required :  To  construct  a  triangle  having  sides  equal  to  A,  B,  C, 
respectively. 

As  base,  draw  a  line  BE  =  A  say. 

From  B  as  center,  with  radius  equal  to  B,  draw  an  arc. 
From  E  as  center,  with  radius  equal  to  C,  draw  another  arc. 
Since  BF  -{-FE>  BE,  the  central  distance,   (Hyp.) 
the  arcs  will  intersect  in  some  point  F. 
'     Join  BF,  EF.     BEF  is  the  required  A ; 
its  sides  being  equal  to  A,  B,  C,  resp.      q.e.f.     (Const.) 

Scholium.  If  the  given  lines  are  all  equal,  the  triangle 
will  be  equilateral;  if  two  are  equal,  the  triangle  will  be 
isosceles. 

Exercise  216.  The  angle  formed  by  two  tangents  drawn  to  a 
circle  from  the  same  point,  is  supplementary  to  that  formed  by  the 
radii  to  the  points  of  contact. 

217.  Show  that  the  angle  formed  by  those  radii  is  bisected  by  the 
line  joining  its  vertex  with  the  center  of  the  circle. 

218.  If  a  tangent  and  a  secant  be  drawn  to  a  circle  from  the  same 
point,  the  angle  formed  by  them  will  be  equal  to,  or  supplementary  to, 
that  formed  by  radii  perpendicular  to  them,  according  as  they  do  or 
do  not  lie  on  the  same  side  of  the  center. 


CONS  TR  UC  Tl  ONS.  99 

Proposition  XVIII.     Problem. 

206.  To  construct  a  parallelograin  with  adjacent 
sides  respectively  equal  to  two  given  lines,  and  includ- 
ing a  given  angle. 


Given :        Two  straight  lines  A  and  B,  and  an  angle  C ; 
Required :  To  construct  a  parallelogram  with  two  sides  equal 
to  A,  B,  respectively,  including  an  angle  C. 

Draw  a  line  DE  =  A,  and  at  D  make  Z  GDE  =  Zc.    (203) 

On  DG  lay  off  DF  =  B. 
From  F  as  center,  with  radius  equal  to  DE,  describe  an 

arc  HK. 
From  E  as  center,  with  radius  equal  to  DF,  describe  an 
arc  HL. 

Let  the  arcs  intersect  in  H,  and  jpin  FH,  EH; 
DII  is  the  parallelogram  required. 
Since  FII  =  DE,  EH  =  DF,  and  Zd  =  Zc,   (Const.) 
DH  is  a  parallelogram    (141),   with  sides  =  A,  B,  resp., 
and  with  an  angle  =  Z  C.  q.e.f. 

Scholium.  When  the  given  angle  is  a  right  angle,  this 
construction  gives  a  rectangle.  When  the  given  sides  are 
equal,  the  construction  gives  a  square  or  rhombus,  according 
as  the  given  angle  is  or  is  not  a  right  angle. 

Exercise  219.  Construct  a  rectangle  having  one  side  double  the 
other. 

220.   Construct  a  square  and  circumscribe  a  circle  about  it, 


100  PLANE   GEOMETRY.  — BOOK  11. 

Proposition  XIX.     Problem. 

207.  To  divide  a  given  straight  line  into  any  num- 
ber of  equal  parts. 

lu      IS      n     h  10 .9 d... 


Given :  A  straight  line  AB  ; 

Required :      To  divide  AB  into  n  equal  parts. 

Let  n  =  l.     Draw  AC,  making  any  convenient  Z  with  AB^ 

and  draw  BD  II  to  AC.  (204) 

On  A  C  lay  off  7  equal  parts,  A-1,  •  •  •  6-7,  of  any  convenient 

length. 
On  BD  lay  off  7  equal  parts,  ^-8,  ••=  13-14,  of  the  same 
length  as  in  AC. 
Join  ^-14,  1-13,  •••  1-B.    AB  is  divided  into  7  equal  parts. 
Since  A-1  is  equal  and  parallel  to  13-14,    (Const.) 
^-14  is  II  to  1-13.  (142) 

Similarly,  all  the  lines  2-12,  3-11,  ••.  1-B,  are  ||  to  ^-14. 
Since  these  II 's  cut  off  7  equal  intercepts  on  A-1,  (Const.) 
these  ll's  cut  off  7  equal  intercepts  on  AB,       (151) 
i.e.,  AB  is  divided  into  7  equal  parts.  q.e.f. 

We  proceed  in  the  same  way  if  n  is  any  other  integer. 

Scholium.  In  practice,  it  will  be  sufficient  to  lay  off  only 
one  part  on  BD,  as  -B-8 ;  then  by  joining  8  with  the  corre- 
sponding division  on  AC,  as  6,  we  cut  off  one  of  the  required 
equal  parts  on  AB  ;  then  lay  off  the  others. 


CONSTRUCTIONS.  101 

208.  Definition.  A  circle  is  said  to  circumscribe  sl 
polygon  when  its  circumference  passes  through  each  of  the 
angular  points  of  the  polygon,  which  is  then  said  to  be 
inscribed  in  the  circle. 

209.  Definition.  A  circle  is  said  to  be  inscribed  in  a 
polygon  when  it  touches  each  of  the  sides  of  the  polygon, 
which  is  then  said  to  be  circumscribed  about  the  circle. 


Proposition  XX.     Problem. 

210.  To  circuTYhscrihe  a  circle  about,  or  to  inscribe 
a  circle  in,  a  given  triangle. 


Given :  Two  triangles  AB  C,  A 'B  'c'. 

Required:  1°,  To  circumscribe  a  circle  about  ABC; 
2°,  To  inscribe  a  circle  m  a'b^c\ 

1°.  When,  by  the  construction  given  in  Prop.  XI.,  a  cir- 
cumference has  been  passed  through  the  three  points  A,  B,  C, 
we  have  a  circle  circumscribing  A  ABC  (208).  q.e.f. 

2°.  When,  by  the  construction  given  in  Art.  153,  we  have 
found  a  point  o'  that  is  equidistant  from  A'b',  A'c',  B'c', 
then  the  circle  described  from  o'  as  center,  with  any  of  the 
equal  perpendiculars  from  O'  to  the  sides  as  radius,  will 
touch  each  of  the  sides,  and  thus  be  inscribed  in  A  a'b'c' 

(209).  *Q.E.F. 


102  PLANE   GEOMETRY.  — BOOK  II. 

PLANE  LOCI. 

211.  The  nature  of  loci  becomes  clear  when  we  conceive 
a  line  as  being  the  path  generated  by  a  point  moving  ac- 
cording to  some  specified  condition.  If  the  point  move  so 
as  to  be  always  at  a  given  distance  from  a  given  fixed  point, 
the  line  thus  traced  will  be  the  locus  of  all  points  that  are 
at  the  given  distance  from  the  given  point.     Hence, 

A  locus  in  a  plane  is  the  line,  or  lines,  every  point  in 
which  satisfies  certain  conditions  fulfilled  by  no  point  not 
in  that  line  or  lines. 

The  following  theorems  upon  the  subject  are  merely  the  enuncia- 
tion from  a  special  point  of  view  of  certain  theorems  already  proved. 
Though  the  proofs  are  here  merely  indicated  by  references,  the  student 
would  find  it  useful  to  frame  a  demonstration  of  each  of  these  theo- 
rems, keeping  in  mind  that,  in  order  that  a  line  be  a  locus, 

(1)  Evey-y  point  in  the  line  must  satisfy  the  given  conditions; 

(2)  No  point  not  in  the  line  should  satisfy  them. 

212.  Theorem.  The  locus  of  all  the  points  situated  at  a 
given  distance  from  a  given  point  is  the  circumference 
described  from  that  point  as  center  with  a  radius  equal  to  the 
given  distance.     (162) 

213.  Theorem.  The  locus  of  all  the  points  that  are  equi- 
distant from  two  given  points  is  the  p)erpendicular  at  the 
mid  point  of  the  line  joining  those  points.     (20,  96) 

214.  Theorem.  The  locus  of  all  the  points  situated  at  a 
given  distance  from  a  given  straight  line  consists  of  the 
two  parallels  drawn  at  the  given  distance  from  the  given  line, 
one  on  each  side.     (139) 

215.  Theorem.  The  locus  of  all  the  points  thai  are  equi- 
distant from  two  given  parallels  is  the  parallel  through  a  point 
equidistant  from  both.     (139) 

216.  Theorem.  The  locus  of  all  the  points  that  are  equi- 
distant from  two  intersecting  lines  consists  of  the  bisectors 
of  the  vertical  angles  formed  by  those  lines.     (101) 


EXERCISES.  103 

217.  Theorem.  The  locus  of  the  mid  points  of  parallel 
chords  is  the  diameter  perpendicidar  to  those 'chords.     (172) 

218.  Theorem.  The  locus  of  the  mid  points  of  equal  chords 
is  the  circumference  conce7itric  with  the  given  circumference, 
and  having  a  radius  equal  to  the  common  distance  of  the 
chords.     (182) 

Familiarity  with  these  and  similar  theorems  is  of  great 
imj)drtaiice,  especially  in  the  solution  of  problems,  which 
very  often  depends  upon  the  intersection  of  loci.  The  fol- 
lowing easy  exercises  will  be  found  useful  as  an  introduction 
to  the  application  of  loci  to  the  solution  of  problems. 


EXERCISES. 

LOCI. 


Note.  — The  line  AB  referred  to  in  these  exercises  is  supposed  to  be  a 
straight  line  of  indefinite  —  that  is,  of  any  requisite  —  length. 

221.  In  the  line  AB  find  a  point  that  shall  be  at  a  given  distance 
CB  from  a  given  point  P.  State  the  conditions  under  which  no  such 
point  can  be  found  ;  one  point ;  more  than  one. 

222.  In  the  line  AB  find  a  point  that  shall  be  equidistant  from  two 
given  points  Pand  Q.     State  the  conditions,  etc.,  as  in  Exercise  221. 

223.  In  the  line  AB  find  a  point  that  shall  be  at  a  given  distance 
CD  from  a  given  line  EF.     State  the  conditions,  etc. 

224.  In  the  line  AB  find  a  point  that  shall  be  equidistant  from  two 
given  parallels  CD,  EF.    State  the  conditions,  etc, 

225.  In  the  line  AB  find  a  point  that  shall  be  equidistant  from  two 
intersecting  lines  CD,  EF.     State  the  conditions,  etc. 

226.  In  a  given  circumference  ABM  find  a  point  that  shall  be  at  a 
given  distance  CD  from  a  given  point  P.    State  the  conditions,  etc. 

227.  In  a  given  circumference  ABM  find  a  point  that  shall  be  equi- 
distant from  two  given  points  P  and  Q.     State  the  conditions,  etc. 

228.  In  a  given  circumference  ABM  find  a  point  that  shall  be  at  a 
given  distance  CD  from  a  given  line  EF.     State  the  conditions,  etc. 


104 


PLANE   GEOMETRY.  — BOOK  II. 


229.  In  a  given  circumference  ABM  find  a  point  that  shall  be  equi- 
distant from  two  given  parallels  CD,  EF.     State  the  conditions,  etc. 

230.  In  a  given  circumference  AB3Ifm6.  a  point  that  shall  be  equi- 
distant from  two  intersecting  lines  CD,  EF.    State  the  conditions,  etc. 

231.  Find  a  point  equidistant  from  two  given  points  Pand  Q,  and 
at  a  given  distance  CD  from  a  given  line  AB,    State  the  conditions,  etc. 

232.  Find  a  point  equidistant  from  two  given  points  P  and  ^,  and 
also  equidistant  from  two  parallels  AB  and  CD.  State  the  con- 
ditions, etc. 

,233.  Find  a  point  equidistant  from  two  given  points  P  and  ^,  and 
also  equidistant  from  two  intersecting  lines  AB  and  CD.  State  the 
conditions,  etc. 

234.  Within  a  given  angle  BAC  find  the  point  that  is  at  a  given 
distance  DE  from  each  of  the  sides. 

235.  Find  a  point  that  shall  be  at  a 
given  distance  AB  from  a  given  point 
P,  and  at  a  given  distance  CD  from 
a  given  point  Q.  State  the  con- 
ditions, etc. 

236.  Find  the  locus  of  a  point  P 
such  that  the  sum  of   its  distances 
from  the  sides  of  a  given  angle  BAG 
is    always  equal    to    a 
given  line  DE.    (101) 

237.  Find  the  locus 
of  a  point  P  such  that 
the  difference  of  its 
distances  from  the  sides  •• 
of  a  given  angle  BAC 
is  always  equal  to  a 
given  line  DE.    (101) 

238.  Find  the  locus 
of  the  mid  points  of  all 
the  lines  drawn  from 
0  to  a  line  AB.     (147) 

239.  Find  the  locus 
of  the  mid  point  of  a 

ladder  whose  lower  end  is  being  pulled  away  from  a  wall.     (144) 


ANALYSIS   OF  PROBLEMS.  105 

ANALYSIS    OP    PROBLEMS. 

Most  of  the  foregoing  simple  exercises  are  problems 
depending  so  plainly  for  solution  upon  the  intersection  of 
certain  loci  that  no  previous  analysis  is  necessary.  In 
general,  the  solution  of  a  problem  will  be  found  to  depend 
upon  several  problems  or  theorems  already  solved  or  proved, 
and  the  solution  will  be  greatly  facilitated  by  a  previous 
analysis.  From  the  nature  of  the  case,  no  precise  rules  can 
be  given,  but  the  general  course  of  procedure  is  expressed 
by  the  following  rules. 

1.  Construct  a  diagram  in  accordance  tvith  the  statement 
of  the  problem,  as  if  the  required  construction  ivere  effected. 

2.  Study  the  relations  of  the  lines,  angles,  etc.,  in  the  dia- 
gram, so  as  to  discover  whether  the  assumed  solution  can  be 
made  to  depend  upon  some  known  problem  or  theorem,  espe- 
cially those  concerning  loci. 

3.  If  such  dependence  cannot  be  found  by  means  of  the 
original  diagram,  make  such  additions  to  it  as  the  case  may 
suggest,  by  joining  points,  draiving  parallels  or  perpendiculars, 
etc.,  and  proceed  as  in  2.    . 

4.  On  discovering  the  dependence  of  the  solution  upon  some 
known  theorem  or  problem,  make  this  the  basis  of  a  synthetic 
solution,  proceeding  in  reverse  order  through  the  several  steps 
of  the  previous  analysis,  till  it  is  shown  that  the  problem  is 
solved.     For  example : 

219.    Problem.    Through  a  given  point  within  an  angle 
draw  a  straight  line  intercepted  between 
the  sides  and  bisected  in  the  given  point.  ^X 

Analysis.   1.    Let  P  be  the  given  yy 

point  within  Z  BAC,  and  suppose  xy  1)^^/ 

drawn  through  P  so  that  Px  =  Py.  y^     7 

2.   The  diagram  as  it  stands  is  not      y"^        / 

suggestive.  Ay  b 


106  PLANE    GEOMETRY.— BOOK  11. 

3.  Through  P  draw  PD  II  to  AB,  to  meet  AC  in  D. 

4.  It  is  at  once  apparent  that  since  PD  is  drawn  through 

the  mid  point  of  xy  and  is  II  to  AB, 

.'.  Z)  is  the  mid  point  of  Ax.  (1^7) 

Synthesis.  Having  thus  found  that  the  solution  depends 
upon  Prop.  XLI.  (147),  we  proceed  as  follows : 

Through  P  draw  PD  II  to  AB  to  meet  AC  in  D.     (109) 
In  DC  take  Dx  =  DA,  and  through  x  draw  xPy,  meeting 
AB  in  y.     xy  is  the  required  line. 

Since  D  is  the  mid  point  of  Ax,         ")         /n      i.\ 

and  DP  is  II  to  AB,  the  base  of  AxAy,  )         ^  "^ 

.-.  xy  is  bisected  in  P.  q.e.f.      (147) 

The  complete  diagram  required  in  the  solution  of  a  prob- 
lem contains : 

1.  The  data,  or  given  figure ;  as  point  P  in  Z  BAC,  above. 

2.  The  qiimsita,  or  things  required ;  as  the  line  xPy. 

3.  Construction  lines  employed  as  auxiliaries  in  the  solu- 
tion ;  as  PD. 

Of  these,  the  data  may  conveniently  be  distinguished,  as 
above,  by  the  first  letters  of  the  -alphabet ;  the  qusesita,  as 
far  as  requiring  new  letters,  by  the  last  letters;  and  the 
auxiliary  constructions  by  dotted  lines. 

We  give  another  example  in  order  to  show  the  application 
of  the  principle  of  intersecting  loci  to  the  solution  of 
problems. 

Pkoblem.    Describe  a  circumference  passing  through  two 
given  points  and  having  its  center 
in  a  given  straight  line. 

Analysis.  1.  Let  AB  be  the 
given  line,  and  C,  D,  the  given 
points ;  and  suppose  circle  CDE, 
having  its  center  at  X  in  AB, 
passes  through  c  and  D. 


nXEkCtSES. 


107 


2.  Since  X  is  equidistant  from  c  and  i),  it  should  be 
found  in  the  locus  of  all  points  that  are  equidistant  from  C 
and  D.  Now  that  locus  is  (213)  the'  perpendicular  at  the 
mid  point  of  the  line  joining  C  and  D  ; 

.:  X  is  found  at  the  intersection  of  that  locus  with  AB. 

Synthesis.  Join  en.  Draw  the  J-FH  at  the  mid  point 
of  CD,  and  let  FH  intersect  AB  in  X.  From  X  as  center, 
with  a  radius  equal  to  the  distance  XC,  describe  the  circum- 
ference CDE.     CDE  is  the  required  circumference. 

Since  X  is  a  point  in  the  ±  at  the  mid  point  of  CD, 

X  is  equidistant  from  C  and  D,  (213) 

.-.  a  circumference   passing   through    C   will   also    pass 

through  D,  (164) 

.*.  a  circumference  having   its   center   in   AB  has   been 

described  through  C  and  D.  q.e.f. 

Scholium.  The  problem  becomes  impossible  in  a  certain 
case.     In  what  case  ? 


EXERCISES 

PROBLEMS. 


240.  Construct  an  isosceles  triangle  having  its 
sides  each  double  the  length  of  the  base. 

241.  Upon  a  given  base  AB,  construct  a  right 
isosceles  triangle. 

242.  With  a  given  line  ^S  as  diagonal,  con- 
struct a  square. 

24.3.    Construct  an    equilateral    triangle    having 
a  given  altitude  AB. 


108 


PLANE   GEOMETRY.  — BOOK  IT. 


244.  In  any  side  of  a  triangle,  find 
the  point  which  is  equidistant  from  the 
other  two  sides. 

245.  In  any  side  of  a  triangle,  find 
the  point  from  which  the  lines  drawn 
parallel  to  the  other  side  are  equal. 

246.  Trisect  a  given  right  angle. 

247.  In  a  given  line  AB^  find  a  point 
X  such  that  the  angle  formed  by  lines 
drawn  from  X  to  two  given  points  C, 
2),  on  opposite  sides  of  AB^  shall  be 
bisected  by  AB. 

248.  In  a  given  line  AB.,  find  a  point 
X  such  that  the  angles  formed  by  lines 
drawn  from  Xto  two  given  points  C,  2), 
on  the  same  side  of  AB,  shall  be  equal. 
How  may  Exercises  247  and  248  be 
enunciated  as  one  problem  ? 

249.  From  a  point  P,  without  a  given 
line  AB,  draw  a  line  PXsuch  that  BXA 
shall  equal  a  given  angle  C 

250.  Find  the  bisector  of  the  angle 
that  would  be  formed  by  two  given  lines, 
without  producing  the  lines. 

251.  Through  a  given  point  P,  draw 
a  line  that  cuts  off  equal  parts  from  two 
intersecting  lines. 

252.  Draw  an  intercept  parallel  to 
the  base  of  a  triangle  such  that  it  shall 
be  equal  to  the  sum  of  the  intercepts 
between  it  and  the  base. 

253.  From  a  given  isosceles  triangle 
cut  off  a  trapezoid  having  for  base  that 
of  the  triangle,  and  having  its  other 
three  sides  equal. 


1 


EXERCISES. 


109 


254.  Three  lines  being  given  diverging 
from  a  point,  draw  a  fourth  line  cutting 
them  so  that  the  intercepted  segments  shall 
be  equal. 

255.  Construct  an  isosceles  right 
triangle,  the  sum  of  the  hypotenuse 
and  a  side  being  given. 

256.  Construct  an  isosceles  right 
triangle,  the  difference  of  the  hypot- 
enuse and  a  side  being  given. 


257. >  Two  angles  of  a  triangle  being  given, 
find  the  third  angle. 

258.  Construct  an  isosceles  triangle  of 
given  altitude,  whose  sides  pass  through  two 
given  points,  and  whose  vertex  is  in  a  given 
straight  line. 

Construct  an  isosceles  triangle,  having 
given : 

259.  The  base  and  the  vertical  angle. 

260.  The  base  and  a  base  angle. 

261.  An  arm  and  the  vertical  angle. 
Construct  a  triangle,  having  given  : 

262.  Two  sides  and  the  included  angle. 

263.  The  base  and  the  base  angles. 

264.  The  three  sides,  AB,  AC,  BC,  such  that  AC  =  ^AB,  and 
BC  =  iAC. 

265.  Construct  an  isosceles  right  triangle,  having  given  the  sum 
and  the  difference  of  the  hypotenuse  and  an  arm. 

Hint.  —  It  is  useful  to  remember  that  ^  and  B,  being  any  two  mag- 
nitudes, 

(A  +  B)-{-iA-B)=2A;  (A-^  B)  -  {A- B)  =2B. 

266.  Construct  a  right  triangle,  having  given  an  arm  and  the  alti- 
tude from  the  right  angle  upon  the  hypotenuse. 

267.  Construct  a  right  triangle,  having  given  the  hypotenuse  and 
the  difference  of  the  other  sides. 


110  PLANE    GEOMETRY.  — BOOK   11. 

Construct  a  parallelogram,  having  given : 

268.  Two  adjacent  sides  and  a  diagonal. 

269.  A  side  and  both  diagonals. 

270.  Both  diagonals  and  their  included  angle. 
Construct  a  trapezoid,  having  given : 

271.  The  four  sides. 

272.  The  parallel  sides  and  the  diagonals. 

273.  The  parallel  sides,  a  diagonal,  and  the  angle  formed  by  the 
diagonals. 

274.  Through  a  point  within  a  circle,  draw  a  chord  that  is  bisected 
in  that  point,  and  show  it  is  the  least  chord  through  that  point. 

275.  The  position  and  magnitude  of  two  chords  of  a  circle  being 
given,  describe  the  circle. 

276.  In  a  given  circle,  draw  a  chord  whose  length  is  double  its 
distance  from  the  center. 

277.  Draw  that  diameter  of  a  given  circle,  which,  being  produced, 
meets  a  given  line  at  a  given  distance  from  the  center.  When  is  this 
impossible  ? 

278.  Describe  a  circle  with  given  radius,  to  touch  a  given  line  in  a 
given  point.     How  many  such  circles  can  be  described  ? 

279.  Describe  a  circle  of  given  radius  to  touch  two  intersecting 
lines.     How  many  such  circles  can  be  described  ? 

280.  Describe  a  circle  touching  two  intersecting  lines  at  a  given 
distance  from  their  intersection.  How  many  such  circles  can  be 
described  ? 

281.  Describe  a  circumference  passing  through  a  given  point,  and 
touching  a  given  line  in  a  given  point. 

282.  Describe  a  circumference  touching  two  given  lines,  and  passing 
through  two  given  points  between  those  lines. 

283.  From  a  given  center,  describe  a  circumference  that  bisects  a 
given  circumference. 

284.  With  a  given  radius,  describe  a  circle  touching  two  given 
circles. 


Book  HI. 

RATIO.     PROPORTION.     LIMITS. 

MEASUREMENT. 

For  many  purposes,  as  in  the  propositions  thus  far  con- 
sidered, it  is  sufficient  to  prove,  in  regard  to  two  given 
magnitudes,  that  they  are  equal  or  unequal.  Thus  we 
proved,  in  Prop.  XXIV.  (99),  that  PA  =  PB,  and  PC>  PA. 
We  have  now  to  consider  how  to  proceed  when  we  wish  to 
estimate  exactly  the  relative  greatness  of  given  magnitudes. 

220.  To  measure  a  magnitude  is  to  find  out  how  many 
times  it  contains  another  magnitude  of  the  same  kind. 

Thus  we  measure  a  line  by  finding  how  many  times  ic 
contains  another  line  called  the  U7iit  of  length,  or  linear  unit. 
This  unit  may  be  either  a  standard  unit,  as  an  inch,  a  meter, 
etc.,  or  a  unit  found  by  dividing  a  line  into  any  desired  num- 
ber of  equal  parts,  as  in  Prop.  XIX.  (207). 

221.  A  quantity  is  a  magnitude  conceived  as  consisting  of 
some  number  of  equal  parts. 

Thus  AB,  regarded  merely  as  a  line,  is 
a  magnitude ;  but  when  thought  of,  or  re- 
ferred to,  as  22  millimeters,  or  x  linear 
units  of  any  kind,  it  is  a  quantity  meas- 
ured by  millimeters,  or  some  other  unit.  The  angle  BAO, 
again,  if  referred  to  as  an  angle  of  31°  15'  47.2",  is  a  quan- 
tity measured  by  tenths  of  seconds. 

Ill 


112  PLANE   GEOMETRY.  — BOOK  III. 

222.    The  number  that  expresses  how  many  times  a  quan- 
tity contains  the  unit  is  called  the  nmnerical  measure  of 
that  quantity.     Ths  numerical  meas- 
ure may   be   a  number   of  any  kind,         -^— - — ■ r 


integral  or  fractional,  rational  or  irrar     ^ b 

tional,  or  any  letter  denoting  number. 

Thus  \i  AB  is  divided  into  11  equal  parts,  of  which  CD 
is  found  to  contain  8|,  and  EF,  x  parts,  then  11,  8|,  and  x 
are  the  numerical  measures  of  AB,  CD,  and  EF,  respectively. 

If  two  quantities  of  the  same  kind,  but  expressed  in  dif- 
ferent units,  are  to  be  compared  as  quantities,  it  is  evident 
that  both  must  be  expressed  as  consisting  of  units  of  the 
same  kind  before  the  comparison  can  be  effected. 

Thus  if  AB  were  8  rods  and  CD  15  yards,  we  cannot 
compare  them  until  we  have  expressed  them  as  132  feet  and 
45  feet,  for  example.  In  every  case,  then,  where  numerical 
measures  are  compared,  they  are  to  be  understood  as  refer- 
ring to  the  same  unit.  It  may  also  be  observed  that  though 
abstract  numbers  are  properly  only  the  numerical  measures 
of  quantities,  yet  they  are  conveniently  regarded  as  quanti- 
ties whose  unit  is  not  expressed. 

223.  If  a  quantity  is  contained  an  exact  number  of  times 
in  each  of  two  like  quantities,  it  is  called  a  common  measure 
of  these  quantities,  which  are  then  said  to  be  commensurable 
with  each  other. 

Thus  a  rod  and  a  yard  are  commensurable,  since  they 
have  2,  3,  6,  9  inches  as  common  measures.  1\  inches  and 
13J^  inches  are  commensurable,  since  they  have  f  of  an  inch 
as  a  common  measure ;  and  any  two  quantities  are  commen- 
surable when,  referred  to  the  same  unit,  their  numerical 
measures  are  rational  numbers. 

224.  If  two  quantities  have  no  common  measure,  they 
are  said  to  be  incommensurable  with  each  other. 

Thus  every  unit  of  our  common  system  of  measures  is 


RATIO.  113 

incommensurable  with  every  like  unit  of  the  metric  sys- 
tem ;  and,  as  afterwards  will  be  seen,  the  diagonal  and  side 
of  a  square  are  incommensurable  with  each  other,  as  are 
also  the  diameter  and  circumference  of  a  circle,  etc.  When 
such  quantities  are  compared,  their  numerical  measures  are 
either  approximate,  as  when  we  say  3  meters  =  118.11237  -|- 
inches,  or  the  numerical  measures  are  such  symbols  as  y'2, 
-^3,  TT,  etc. 

RATIO. 

225.  The  ratio  of  two  quantities  is  their  relative  greatness 
as  expressed  by  the  quotient  of  the  one  by  the  other. 

Thus  the  ratio  of  15  inches  to  7  inches  is  15  in.  -j-  7  in.  =  \^- ; 
the  ratio  of  8  rods  to  15  yards  is  (reducing  to  a  common 
measure)  132  ft.  -j-  45  ft.  =  ^^-^- ;  and,  generally,  if  the  nu- 
merical measures  of  two  quantities  are  a  and  h,  the  ratio  of 

these  quantities  is  -,  a  and  b  being  any  numbers  whatever. 

226.  The  ratio  of  the  quantity  ^  to  a  like  quantity  B  is 
denoted  symbolically  by  either  of  the  expressions 

A  :  B,  or    — , 
B 

each  of  which  is  read,  the  ratio  of  A  to  B.  In  each,  A  is 
called  the  first  term  or  antecedent;  B,  the  second  term  or 
consequent. 

227.  When  A  and  B  are  commensurable  quantities,  the 
value  of  their  ratio  is  expressed  exactly  by  the  fraction  de- 
noting the  quotient  of  the  numerical  measure  of  the  ante- 
cedent by  that  of  the  consequent.  But  if  the  quantities 
are  incommensurable,  no  rational  fraction  can  express  their 
ratio  exactly,  since  then  they  would  not  be  incommensurable. 
Yet,  by  taking  the  unit  of  measure  sufficiently  small,  we  can 
find  a  fraction  that  expresses  the  true  value  of  the  ratio  to 
as  near  a  degree  of  approximation  as  we  please. 

Geom.  —  8 


114  PLANE   GEOMETRY.  — BOOK  III. 

Thus  suppose  we  have  two  lines  A  and  B,  whose  nu- 
merical measures  are  -yj2  and  1,  re- 
spectively.    Now,  carried  to  seven 

decimals,  ^2  =  1.4142135  +  ;  that      ^ 

is,  V2  >  1.414213  and  <  1.414214,  so  that  the  ratio  of  A  to 
B  or  ^2 : 1  lies  between 

TFoiroinj"  ^^^  1 0  0  0  0  o'O"? 

and  must  differ  from  either  by  less  than  one  millionth.  As 
we  can  find  the  value  of  -^2  to  any  number  of  decimals,  we 
can  find  a  fraction  that  differs  from  ^2 : 1  by  less  than  any 
assignable  quantity. 

To  generalize,  let  A  and  B  be  any  two  incommensurable 
quantities.  If  we  suppose  B  divided  into  any  number  of 
equal  parts,  so  that  B  =  nP,  P  denoting  one  of  the  parts, 
then  A  must  contain  some  number  m  of  such  parts,  with  a 
remainder  less  than  P  ;  or  ^  >  m  P  and  <  (m  -}-  1)P.     Thus 

B      nP  nP 

*  A  'Yfh  T/h       I       1 

that  is,  —  lies  between  —  and  — -^--,  and  must  differ  from 
B  n  n     1 

either  of  these  fractions  by  less  than  -,  a  fraction  that  may 

be  made  less  than  any  assigned  quantity  by  taking  n  suffi- 
ciently great.     Hence 

A. rational  fraction  can  he  found  that  expresses  the  ratio  of 
any  two  given  incommensurable  quantities  within  any  required 
degree  of  x>recision. 

PROPORTION.    DEFINITIONS. 

228.  If  two  pairs  of  quantities  have  equal  ratios,  they 
are  said  to  be  proportionals  or  to  be  in  proportion.     Thus 

A       G 
each  of  the  equalities  A:  B  —  C:D,  —  =  — , 

B      D 


PROPORTION.  115 

expresses  a  proportion  that  may  be  read,  the  ratio  of  A  to  B 
is  equal  to  the  ratio  of  C  to  D  ;  or  more  briefly,  A  is  to  B  as 
C  is  to  D. 

229.  The  first  and  fourth  terms  of  a  proportion  are  called 
its  extremes;  the  second  and  third,  its  means.  The  fourth 
term  is  also  called  a  fourth  proportional  to  the  other  three. 

230.  Although  ratio,  from  its  very  nature,  can  exist 
between  like  quantities  only,  yet  as  we  may  have,  for 
example,  Z.  A~  ^/.B,  and  also  arcP  =  |  arc  Q,  the  pro- 
portion 

Z.  A  :  Zb  =  arc  P  :  arc  Q, 

may  be  stated  between  these  pairs  of  unlike  quantities, 
since  the  proportion  simply  states  that  the  angle  A  is  just 
as  great  compared  with  the  angle  B  as  is  the  arc  P  compared 
with  the  arc  Q. 

231.  Of  the  following  theorems  concerning  proportions 
and  their  transformations,  some  apply  to  pairs  of  quantities 
whether  like  or  unlike,  and  the  given  proportion  will  be 
stated  under  the  form 

A:B  =P:Q. 

Others  apply  only  when  the  pairs  of  quantities  are  like,  and 
the  given  proportions  will  be  stated  under  the  form 

A:B  =  C:B. 

Others,  again,  that  apply  properly  to  numbers  only,  will 
have  the  given  proportion  stated  under  the  form 

a:b  =  c:d. 

232.  It  follows  at  once,  from  the  definitions  of  quantity, 
ratio,  and  propoj'tion,  that  (1)   if  four  quantities  are  in  pro- 
portion, their  numerical  measures  are  also  in  proportio7i ;  i.e. 
if  A  :  B  =  P  :  Q, 

then  a:h  =  p:q,  or  -  = -, 

^  'h      q 


116  PLANE   GEOMETRY.  —  BOOK  III. 

and  conversely ;  (2)  if  four  numbers  are  in  proportion,  quan- 
tities of  which  these  yiumhers  are  numerical  measures  are  also 
inproporiion;  i.e., 

.„  .  a     p 

if  a:b  =  p:q,  OY  t  =  -? 

b      q 

then  A.B  =  P:  Q. 

It  also  follows  from  the  same  definitions,  and  Ax.  1,  that 
(3)  ratios  equal  to  the  same  ratio  are  equal  to  each  other. 


EXERCISES. 

QUESTIONS. 


285.  Taking  the  inch  as  unit,  what  is  the  ratio  of  1  ft.  to  7  in.  ?  To 
13in.  ?     Tol^ft.  ?    To  21  ft.  ?    To  |  yd.  ? 

286.  A  train  goes  at  the  rate  of  112  miles  in  3|  hrs. ;  a  second  train 
at  the  rate  of  105  miles  in  2\  hrs.  What  is  the  ratio  of  the  speed  of 
the  first  train  to  that  of  the  second  ? 

287.  What  is  the  ratio  of  1  lb.  to  9  oz.  ?  To  33  oz.  ?  To  2|  lbs.  ? 
To  20|  lbs.  ? 

288.  That  216  grs.  of  silver  may  be  worth  13|  grs.  of  gold,  what 
should  be  the  ratio  of  the  value  of  gold  to  that  of  silver  ? 

280.  180°  F.  =  100°  C.  =  80°  R.  What  is  the  ratio  of  1°  F.  to  1°  C. 
and  1  R.°,  respectively  ? 

290.  The  vertical  angle  of  an  isosceles  triangle  is  50°.  What  is  the 
ratio  of  that  angle,  (1)  to  a  right  angle  ?  (2)  to  each  of  the  base 
angles  ? 

291.  A  base  angle  of  an  isosceles  triangle  is  75°.  What  is  the 
ratio  of  that  angle,  (1)  to  a  right  angle  ?  (2)  to  the  vertical  angle  ? 

292.  The  ratio  of  a  base  angle  of  an  isosceles  triangle  to  the  vertical 
angle  is  f .     What  is  that  angle,  and  what  is  its  ratio  to  a  right  angle  ? 

293.  An  acute  angle  of  a  right  triangle  is  35°.  What  is  the  ratio  of 
that  angle  to  the  other  acute  angle  ? 

294.  A  certain  angle  has  to  an  angle  of  an  equilateral  triangle  the 
same  ratio  that  the  latter  has  to  a  right  angle.  How  many  degrees  in 
the  first  angle  ? 


LIMITS.  117 

LIMITS. 

Thus  far  the  magnitudes  considered  have  been  of  fixed 
greatness  only.  In  Art.  227  we  had  occasion,  however,  to 
discuss  certain  approximate  ratios  tending  towards  a  value 
they  can  never  exactly  attain,  though  they  can  approach 
it  indefinitely  near.  In  these  we  have  examples  of  what 
are  termed  a  variable  and  its  limit,  now  about  to  be  defined. 

233.  A  constant  is  a  magnitude  or  quantity  whose  great- 
ness remains  the  same,  neither  iucreasing  nor  decreasing. 

234.  A  variable  is  a  magnitude  or  quantity  whose  great- 
ness goes  on  increasing  or  decreasing. 

Thus  a  chord  .4J5,  as  long  as  it  passes 
through  the  center  O,  remains  constant 
in  every  position.  But  if  it  turn  about 
an  extremity  A,  we  have  a  decreasing 
variable  chord  AC,  Av',  etc.,  while  in 
the  angles  BAG,  etc.,  and  in  the  arcs 
BC,  BC',  etc.,  we  have  increasing  variables. 

235.  If  a  variable  increases  or  decreases  so  as  to  approach 
indefinitely  near  to  an  equality  with  a  certain  constant,  this 
constant  is  called  the  limit  of  the  variable. 

Thus  in  the  figure  above,  the  variable  chord  decreases 
towards  zero  as  limit,  the  variable  angle  increases  towards 
a  right  angle  as  limit,  the  variable  arc  increases  towards  ji 
semicircumference  as  limit,  and  the  distance  of  the  variable 
chord  from  the  center  increases  towards  the  radius  as  limit, 
none  of  which  limits,  evidently,  can  be  attained  as  long  as 
there  is  any  chord. 

Again,  if  a  point  P    o p        p' p"    p'"    ^ 

move   along    from    0 

towards  N,  the  distances  OP  and  PN  are  variables ;  the  one 
increasing  towards  ON,  the  other  decreasing  towards  zero. 
If  no  condition  were  imposed  upon  the  motion  of  P,  it  might 


118  PLANE   GEOMETRY.  — BOOK  111. 

reach  N,  or  even  pass  beyond  it,  and  the  distance  ON  would 
not  be  a  limit  of  OP  according  to  the  definition.  But  if  we 
impose  the  condition  that  at  the  end  of  the  first  second  it 
reach  p',  half  way  to 

N,  at  the  end  of  the    o_ p        p' p"    p'"    n 

next  second  reach  P," 

half  way  between  P'  and  JSf,  and  so  on,  it  is  evident  that 
P  could  never  reach  N,  though  it  might  come  indefinitely 
near  to  it.  For  the  fraction  of  the  distance  passed  over  in 
n  seconds  would  be  the  sum  of  the  series  J  +  -|  +  g-  -j-  yg-  + 
a  series  that  has  1  for  its  limit,  a  limit  the  series  can  never 
attain,  no  matter  how  many  terms  may  be  taken. 


Proposition  I.     Theorem. 

236.  If  two  variables  tending  towards  limits  are 
always  equal,  these  limits  are  also  equal. 

A M^ D      P 

B N Q 

Given :  Two  equal  variables  A  31  and  BN,  tending  towards 
limits  AP  and  PQ; 

To  Prove :  PQ  is  equal  to  ^P. 

For  if  AP  could  be  greater  than  BQ,  some  part  of  AP,  as 
AD,  would  be  equal  to  BQ.  Now,  however  small  the  con- 
stant difference  DP  could  be,  since  A3f  can  increase  so  as  to 
approach  AP  nearer  than  any  constant  difference  (Hyp.), 
A3I  would  become  greater  than  AD  or  its  equal  B  Q,  while 
BN  would  always  remain  less  than  BQ  (Hyp.).  That  is, 
BN  would  be  both  equal  to  and  less  than  AM.  In  the  same 
way  it  can  be  proved  that  B  Q  cannot  be  greater  than  AP. 

Hence  B  Q  must  be  equal  to  AP.     q.e.d. 


THEORY  OF  PROPORTION.  119 

PROPORTION.    THEOREMS. 
Proposition  II.     Theorem. 

237.  If  four  numbers  are  in  proportion,  the  product 
of  the  extremes  is  equal  to  the  product  of  the  means. 

Given:  a:b  =  c:d; 

To.  Prove :  ad  =  be. 

Since  ^  =  ^,  (Hyp.) 

b     d 

.'.  multiplying  both  members  by  bd, 

ad  =  be.  Q.E.D. 

238.  Cor.    If  the  means  are  equal,  that  is,  if 

a  :  b  =  b  :  c, 
then  ac  =  b%  (237) 

i.e.,  the  product  of  the  extremes  is  equal  to  the  square  of  the  mean. 

239.  Definition.  Here  b  is  said  to  be  a  7nean  propor- 
tional between  a  and  c,  and  c  is  a  third  proportional  to  a 
and  b. 

Exercise  295.  If  the  bisector  of  an  angle  of  a  parallelogram  passes 
through  the  opposite  vertex,  the  figure  must  be  equilateral. 

296.  Any  parallelogram  that  can  be  circumscribed  about  a  circle 
must  be  equilateral. 

297.  If  each  of  three  equal  circles  is  tangent  to  the  other  two,  their 
lines  of  centers  form  an  equilateral  triangle. 

298.  In  triangle  ABC,  if  D,  E,  the  mid  points  of  AB,  AC,  be 
joined,  show  that  A  ^5 C:  trapezoid  DBCE  =  4:3. 

299.  If  A  is  an  angle  of  an  equilateral  triangle  and  J5  is  a  right 
angle,  show  that  Z  ^  :  Z  ^  =  2  :  3. 

300.  In  the  diagram  for  Art.  153,  ii  ABAC :  AACB  =  1 '.2>,  and  if 
Z  BAC :  Z  ABC  =7:3,  show  that  /.AOC-.Z.  ABC  =7:2. 

301.  Hence  deduce  the  ratio  of  angle  BOG  to  angle  BAC  and  of 
angle  AOB  to  angle  ACB. 


120  PLANE   GEOMETRY.  — BOOK   III. 

Proposition  III.     Theorem. 

240.  Conversely,  if  the  product  of  two  numbers  is 
equal  to  the  product  of  two  others,  either  two  may 
be  made  the  extremes  of  a  proportion,  and  the  other 
two  its  means. 

Given:  ad  =  bc; 

To  Prove :  a:b  =  c:d. 

Since  ad  =  be,  (Hyp.) 

.-.  dividing  both  members  by  bd, 

9:  — 2. 
b~d! 

i.e.,  a:b  =  c:  d.  q.e.d. 

241.  Scholium.  If,  instead  of  by  bd,  we  divide  by  ab,  ac, 
or  cd,  we  obtain  from  ad  =  be  the  proportions,  each  derivable 
from  any  of  the  others : 

d:b  =  e:  a,  or  c:  a  —  d:b ; 

d  :  c  =  b:  a,  or  b:  a  =  d:  c; 

a  :  e  =  b  :  d,  oy  b:  d  =  a:  c. 

Exercise  302.  In  the  first  diagram  for  Prop.  XIII.  (195),  prove 
that  if  AC  and  BD  be  joined,  angle  BAC  will  be  equal  to  angle  ABD. 

303.  In  the  same  diagram,  show  that  if  AD  and  BC  be  joined, 
they  will  intersect  in  MN. 

304.  In  the  third  diagram  for  the  same  proposition,  if  a  perpen- 
dicular be  drawn  through  the  mid  point  of  EF,  it  will  pass  through 
M  and  iV. 

305.  If  two  tangents  to  the  same  circle  make  equal  angles  with 
an  intercept  between  them  that  passes  through  the  center,  they  are 
equal. 

306.  If  two  secants  to  the  same  circle  make  equal  angles  with  an 
intercept  between  them  that  passes  through  the  center,  they  are  equal. 


THEORY   OF  PROPORTION.  121 

Proposition  IV.     Theorem. 

242.    The  products  of  the  corresponding  terms  of 
two  or  more  numerical  proportions  are  in  proportion. 

Given :  a:b  =  c:  d,  and  e  :f=  g  :  h; 

To  Prove :  ae  :bf=cg  :  dh. 

Since  ^  =  "^,  and  ^  =  f,  (Hyp.) 

h      d  f     h 

.'.  multiplying  member  by  member, 

ae  _  eg 
hf^dJi 

i.e.,  ae :  hf=  eg :  dh.  q.e.d. 

Tn  the  same  way  the  theorem  can  be  proved  for  any 
number  of  proportions. 


Proposition  V.     Theorem. 

243.  If  four  numbers  are  in  proportion,  like  powers 
or  Wee  roots  of  those  numbers  are  in  proportion. 

Given :  a:b  =  c:d; 

i_    1        11 

To  Prove :       a'* :  6"  =  c" :  d'\  and  a"" :  6"  =  c** :  d"". 

Since  ^  =  |,  (Hyp.) 

.'.  taking  like  powers  or  like  roots  of  both  members, 

1^       1 

a"     c"       T  a"      c** 
—  =  —  and  — f  =  — r, 

1       i  1        i 

i.e.,  a" :  b"  =  c" :  d%  and  a" :  6"  =  c" :  d"  q.e.d. 


122  PLANE   GEOMETRY.  — BOOK  III. 

Proposition  VI.     Theorem. 

244.  If  four  like  quantities  are  in  proportion,  they 
are  in  proportion  taken  alternately. 

Given :  A:B  =  C:D; 

To  Prove :  A:C  =  B:D. 

Since  -  =  -,  (Hyp.  and  232') 

h      d 

ad=bc,  (237) 

.-.  a:c  =  b:d,  (241) 

.-.   A:C  =  B:D.  Q.E.D.      (232") 

Scholium.  Alternation,  as  will  be  seen  from  the  symboli- 
cal statement  of  the  theorem,  means  taking  the  alternate 
terms  of  a  given  proportion  so  as  to  form  a  new  one. 

It  is  also  evident  that  this  transformation  can  be  applied 
only  to  a  proportion  in  which  the  terms  are  all  like  terms, 
since,  if  A  and  P  are  unlike  quantities,  no  ratio  can  exist 
between  them. 


Proposition  VII.     Theorem. 

245.  If  any  four  quantities  are  in  proportion,  they 
are  in  proportion  taken  inversely. 

Given:  A:B  =  P  :  Q; 

To  Prove :  B:A  =  Q'.P. 

Since  a:b  =  p:q,  (Hyp.  and  232') 

bjj  =  aq,  (237) 

.-.  b:a  =  q:p,  (241) 

.-.   B:A  =  Q:P.  Q.E.D.      (232") 

Scholium.    Inversion,  as  may  be  seen  from  the  symboli- 
cal statement,  means  taking  the  terms  in  inverse  order. 


THEORY  OF  PROPORTION.  123 

Proposition  VIII.     Theorem. 

246.  //  any  four  quantities  are  in  proportion,  they 
are  in  proportion  taJcen  hy  composition. 


Given : 

A:B  =  P:Q; 

To  Prove : 

A-\-B:B  =  P  -{-  Q:Q. 

Since  ^=^, 
b      q 

(Hyp.  and  232') 

hH^^' 

(Ax.  2) 

••        h      -      q     ' 

.-.   A-\-B'.B  =  P  -\-Q:Q. 

Q.E.D.         (232") 

Similarly, 

A  -\-B:A  =  P  +  Q'.P. 

Proposition  IX.     Theorem. 

247.  If  any  four  quantities  are  in  proportion,  they 
are  in  proportion  taken  hy  division. 


Oiven  : 

A:B  =  P:Q; 

To  Prove: 

A  —  B:B  =  P—Q: 

Q. 

Since  ^  =  ^, 
b      q 

(Hyp.  and  232') 

b              q 

•       (Ax.  3) 

a—b_p—q 

'■       b  g     ' 

.-.  A  -  B  :  B  =  P  -  Q  .  Q.       Q.E.D.      (232") 
Similarly,  A—  B  :A  =  P  ~  Q:  p. 


124  PLANE   GEOMETRY.  — BOOK  III. 

Proposition  X.     Theorem. 

248.  //  any  four  quantities  are  in  proportion,  they 
are  in  proportion  taken  hy  composition  and  division. 
Given :  A:B  =  P'.Q; 

To  Prove :    A-\-B:A  —  b  =  P-{-Q:P  —  q. 

Since  a:h=i^:q,  (Hyp.  and  232') 

a-{-b_p  +  q 

and^-^^  =  ^-^,  (247) 

.*.  dividing  member  by  member, 
a-\-h  _p-{-q 
a  —  b     p  —  q 
,\  A  -{-  B  :  A  —  B  =  P  -\.  Q  :  p  —  Q.      q.e.d.      (232") 


Proposition  XI.     Theorem. 

249.  If  two  proportions  have  the  same  antecedents, 
the  consequents  are  in  proportion,  and  vice  versa. 


Oiven  : 

A:B=P:Q,SindLA 

c  =  P:.r; 

To  Prove: 

B  :C=  Q:B. 

Since  e=.^ 

I     I  \               (Hyp.  and  232') 

and  ^  ^l, 
b     4 

,'.  dividing  member  by  member, 

^  =  % 

c      r 

.'.  B  :C=  Q: 

R.              Q.E.D.     (232") 

250.    Definition.   A  continued  proportion  is  a  series  of 
equal  ratios. 


THEORY  OF  PROPORTION.  125 

Proposition  XII.     Theorem. 

251.  If  any  number  of  like  quantities  are  in  con- 
tinued proportion,  the  sum  of  the  antecedents  is  to 
the  suin  of  the  consequents  as  any  antecedent  is  to 
its  consequent. 

Given:  A:B  =  C:D=E:F; 

To  Prove :  A  -\-  C  -\-  E  :  B  -\-  n  -\-  F  =  A  :  B. 

Since  a:b  =  c:d  =  e:f,  (Hyp.  and  232') 

ab  =  ba,  ad  =  be,  of—  be,  (237) 

...  a(b-^d-^f)=:b(a-{-c-\-e), 

.'.  a-]-c  +  e:bi-d-\-f=a:b,  (240) 

.'.  A-{- C -\-E:B -{-D -{-F  =  A:B.      Q.E.D.      (232") 

In  the  same  way,  the  theorem  may  be  proved  for  any 
number  of  ratios. 


Proposition  XIII.     Theorem. 

252.  If  the  antecedents  of  any  four  quantities  in 
proportion  be  multiplied  by  any  number,  and  the 
consequents  by  any,  the  results  will  be  in  propor- 
tion. 


Given : 

A:B  =  P:Q; 

To  Prove  : 

mA  :nB  =  mP:nQ. 

Since  a  :  b  =  p  :  q, 

(Hyp.  and  232') 

d 

m:n  =  m:  n, 

ma  :  nb  =  mp  :  nq, 

(242) 

,-.  mA  :  nB  =  mP  :  nQ. 

Q.E.D.      (232") 

126  PLANE   GEOMETRY. —BOOK  III. 

Proposition  XIV.     Theorem. 

253.  If  both  terms  of  a  ratio  he  multiplied  or 
divided  by  any  numher,  the  ratio  remains  the  same. 

Given :  Any  ratio  A:  B  ; 

To  Prove :         mA  :  mB  =  A  :  B  =  -A  :-B. 

n       n 

Since  a  :  h  =  a  :  b  =  a  :  b, 

and  m:m  =  l:l  =  -:-, 

n    n 

ma  :mb  =  a:b  =  -a:-b,  (242) 

.'.  mA  :  mB  =  A:B  =  -A:~B.    Q.E.D.  (232") 

254.  Cor.   If  A  .  B  =  p  :  Q, 

then  mA  :  mB  =  nP  :  7iQ,  \ 

11  1       1       i  (253) 

and  —A:—B  =  -P:-Q.\ 

mm         n       n      J 

Exercise  307.  Any  parallelogram  that  can  be  inscribed  in  a  circle 
will  have  the  intersection  of  its  diagonals  at  the  center  of  the  circle. 

308.  Hence,  show  that  rectangles  are  the  only  parallelograms  that 
can  be  inscribed  in  a  circle. 

309.  Any  parallelogram  that  can  be  inscribed  in  a  circle  must  be 
equiangular. 

310.  Describe  a  circumference  passing  through  two  given  points  and 
having  its  center  in  a  given  straight  line.     When  is  this  impossible  ? 

311.  Prove  that  all  circumferences  that  pass  through  a  given  point 
and  have  their  centers  in  a  given  straight  line  must  also  pass  through 
a  second  given  point. 

312.  From  a  given  point  as  center  describe  a  circle  to  touch  a  given 
circle.     How  many  solutions  are  there  ? 


THEORY  OF  PROPORTION.  127 


Proposition  XV.     Theorem. 

255.  Two  incommensurahle  ratios  are  equal  if 
their  corresponding  approximate  values  are  always 
equal. 

A  P 

Given :   Two  ratios  —  and  —  such  that,  when  an  approximate 

value  of  —  is  — ,  the  corresponding  value  of  —  also  is  — ; 

To  Prove :  A  :  B  =  P  :  Q. 

The  supposition  is  that  when  B  and  Q  are  each  divided 
into  n  equal  parts,  if  A  contain  m  parts  of  B,  with  some 
remainder,  then  also  P  contains  m  parts  of  Q,  with  some 
remainder. 

^.         A      m      X       ^  P      m      x'  /XT      \ 

Since  -  =  -  +  -  and -=-  +  -,  (Hyp.) 

B       n       71  Q       n       n 


m  and  n  being  integers,  but  x  and  x'  each  <  1, 

(235) 


—  has  for  limit  —, 
n  B 


also  —  has  for  limit  — , 

71  Q 

X     X 

as  -J  —  each  tend  toward  zero  when  ti  is  indefinitely  great, 

.-.  -  =  -,  (236) 

B      Q 

being  the  limits  of  variables  always  equal, 

i.e.,  A  :  B  =  P  :  Q.  Q.E.D. 

Scholium  1.  It  is  to  be  carefully  noted  that  A  :  B  and 
P  :  Q  are  here  proved  absolutely,  not  app7'oximately,  equal. 

Scholium  2.  In  Props.  II.-XIV.,  we  found  that,  a  pro- 
portion being  given,  certain  transformations  can  be  performed 
on  it.  In  this  proposition  we  have  the  criterion,  or  test,  of 
proportionality  as  regards  both  incommensurable  quantities 
and  those  of  whose  commensurability  we  know  nothing. 


Book  IV. 

PROPORTIONAL  ANGLES  AND  LINES.    SIMILAR 
POLYGONS. 


Xi^f^C 


PROPORTIONAL   ANGLES. 

256.  Definition.  An  angle  formed  by  two  radii  of  a 
circle  is  called  an  angle  at  the  center,  and  is  said  to  intercept 
the  arc  that  lies  between  its  sides. 

Proposition  I.     Theorem. 

257.  In  the  same  circle,  or  in  equal  circles,  equal 
angles  at  the  center  intercept  equal  arcs;  and  con- 
versely. 


1°.  Given:   At  the  centers   of   equal   circles   ADB,   A'd^b', 
angle  O  equal  to  angle  O' ; 

To  Prove:  Arc  ACB  is  equal  to  arc  A'c'b'. 

Join  AB,  A'b'.     Then 
since  OA  =  o'A',  OB  =  o'b',  and  Z  0  =  Zo',     (Hyp.) 

A  GAB  =  A  0' A'B',  (66) 

.-.   AB=A'B',  (70) 

.-.  arc  ACB  =  arc  A'c'b'.  q.e.d.      (174") 
128 


PROPORTIONAL  ANGLES.  129 

2°.  Given:   In   equal  circles  ADB,  a'd'b',  equal  arcs  ACB, 
a'c'b',  intercepted  by  angles  0,  o',  respectively; 
To  Prove:        Angle  0  is  equal  to  angle  o'. 

Join  AB  and  A'b'.     Then 

since  arc  ACB  =  arc  a'c'b',  (Hyp.) 

chord  AB  =  chord  A'b' ;  (174") 

also  OA  =:0'A',  and  OB  =  o'b',  (Hyp.) 

.-.  Zo  =  Zo'.               Q.E.D.  (69,70) 

258.  Cor.  1.  A  radius  bisecting  an  angle  at  the  center 
bisects  its  arc;  and  conversely. 

259.  Cor.  2.  If  angle  A  is  equal  to  m  times  ayigle  B,  then 
the  arc  of  angle  A  is  equal  to  m  times  the  arc  of  angle  B ; 
and  conversely. 

Scholium.  The  conclusion  proved  in  Prop.  I.  might  be 
stated  under  the  form 

Z  0:Z0' =  7)1  :m  =  avG  ACB  :A'c'b', 
such  a  ratio  as  m  :  m  being  called  a  ratio  of  equality. 

Exercise  313.  In  the  same  circle  or  in  equal  circles,  the  greater  of 
two  unequal  angles  at  the  center  intercepts  the  greater  arc. 

314.  An  angle  at  the  center  is  obtuse,  right,  or  acute,  according  as 
its  arc  is  greater  than,  equal  to,  or  less  than  a  quarter  of  a  circumference. 

315.  Intersecting  diameters  intercept  equal  arcs  between  their 
extremities. 

316.  If  the  extremities  of  intersecting  diameters  be  joined,  the 
figure  formed  will  be  a  rectangle. 

317.  In  the  diagram  for  Prop.  I.,  if  angle  0  is  f  of  a  right  angle, 
what  is  the  ratio  of  arc  ACB  to  arc  ADB  ? 

318.  In  the  same  diagram,  if  vlO  be  produced  to  meet  the  circum- 
ference in  E,  what  part  will  arc  BE  be  of  the  circumference,  suppos- 
ing, as  before,  that  angle  0  is  f  of  a  right  angle  ? 

319.  In  the  left-hand  circle  of  the  diagram  for  Prop.  II.,  show  how 
to  bisect  arc  A  CB  without  joining  AB. 

Geom. — 9 


130  PLANE   GEOMETRY.  — BOOK  IV. 

Proposition  II.     Theorem. 

260.  In  the  same  circle,  or  in  equal  circles,  angles 
at  the  center  are  proportional  to  their  intercepted  arcs. 


Given:   In  equal  circles  with  centers  0  and  0\  respectively, 
angles  AOB,  A'o'b',  with  their  respective  arcs  AB,  A^B' ; 
To  Prove :  Angle  AOB  :  angle  ^'o'^'  =  arc  ^5  :  arc  ^'5'. 

1°.    When  the  arcs  AB,  A'b',  are  commensurable. 

Let  B'c'  be  a  common  measure  of  A'b'  and  AB,  so  that 

arc  b'c'  is  contained  5  times  in  A'b'  and  8  times  in 

AB.     Join  O'C'. 

Since  arc  AB  =  S  arc  B'c',  and  arc  A'b' =5  arc  b'c',  (Hyp.) 

Za'ob  =  SZ  b'O'c',  and  Z  a'o'b'  =  5Z  b'o'c',  (259) 

.-.  arc  AB  :  arc  A'b'  =  8:5,] 

^ndZA0B:ZA'0'B'  =  S'.5,\  ^^^^^ 

.'.   ZAOBiZ  A'O'B' =  2iVGAB:^TG  A'B'.    Q.E.D.     (232'") 

2°.   When  the  arcs  AB,  A'b',  are  incommensurable. 
Suppose  A'b'  divided  into  any  number  of  equal  parts  n, 
and  that  AB  contains  this  nth.  part  of  A'b'  m  times, 
with  a  remainder  BC.     Draw  OC. 
Since  AC  and  A'b'  are  commensurable  arcs,    (Const.) 
Zaoc  _'wi_  arc^C 
Z  A'O'B'  ~'n~  SiVG  A'b'' 


(1°) 


Z  AOB       m      X        ,   arc^B       m      x' 

—. — r-7— i  =  — h  -)  and  - —     .    .  = 1 — , 

Z  A'o'b'      n      w  diVcA'B'      n       n 


since  AOB  and  AB  are  slightly  >  AOC  and  AC,  respectively. 


PROPORTIONAL  ANGLES.  131 

Now  when  n  is  taken  indefinitely  great,  -  and  —  become 
indefinitely  small, 

X  and  ic'  being  each  <  1 ; 

/.AOB        arc  ^5  /or--s 

Za'o'b'     sltga'b"  ^      / 

being  the  limits  of  variables  always  equal. 

261.  Definition.  One  quantity  is  said  to  be  measured 
by  another  of  a  different  kind,  if  they  are  so  related  that 
the  numerical  measure  of  the  one  always  expresses  that  of 
the  other  also. 

This  may  also  be  worded  as  follows :  A  quantity  A  is 
measured  by  another  quantity  ^',  if  A  and  its  unit  U  are 
always  in  proportion  to  A'  and  its  unit  U' ; 

i.e.,  ii  A:U  —  A':U'. 

Thus  temperature  is  measured  by  the  number  of  equal 
lengths  in  a  column  of  mercury;  time,  by  the  number  of 
equal  arcs  in  the  circumference  of  a  dial ;  etc. 

262.  Cor.  An  angle  at  the  center  is  measured  by  its  inter- 
cepted arc. 

For  if  A  denote  any  angle,  and  u  its  unit-angle.  A',  the 
intercepted  arc,  and  u'  its  unit-arc,  then  (260), 
ZA:Zu=  arc  A' :  arc  u'  =  m:  1, 
m  being  the  numerical  measure  oi  A'  in  terms  of  its  unit  u'. 

Scholium.  Though  any  convenient  arc  may  be  taken  as 
unit,  that  usually  employed  is  the  90th  part  of  a  quad- 
rant or  i  of  a  circumference,  and  is  called  a  degree.  As  a 
right  Z  is  measured  by  a  quadrant  (262),  the  90th  part  of 
a  right  Z  is  also  called  a  degree.  Each  degree,  again,  is 
divided  into  60  minutes,  and  each  minute  into  60  seconds. 

263.  Definition.  The  angle  formed  by  two  chords  that 
meet  in  the  circumference  is  called  an  inscribed  angle. 


132 


PLANE   GEOMETRY.  — BOOK  IV. 


Proposition  III.     Theorem. 

264.  An  inscribed  angle  is  measured  hy  half  its 
intercepted  arc. 

B, .  B B  ^ 


Given:  An  inscribed  angle ^BC  intercepting  the  arc  AC ; 
To  Prove :  Angle  ABC  i^  measured  by  \  arc  AC. 

Find  o,  the  center  of  the  circle.     Then 

1°.   If  0  lies  in  a  side  BC  oi  /.ABC,  join  AG. 
Since  AO  =  B0, 
Za  =  Zb. 
But  Zaoc=Za-{-Zb  =  2Zb, 

.'.  Z  B  =^ZA0C. 

Now  Z  ^oc  is  meas.  by  arc  AC, 
Z  B  is  meas.  by  ^  arc  AC. 

2°.    If  0  lies  between  BA  and  BC,  draw  BOD,  a  diam. 
Since  Z  ABD  is  meas.  by  |-  arc  AD,  ^ 
and  Z  DBC  is  meas.  by  |-arc  DC,  ) 
Z  ABD  +  Zdbc  is  meas.  by  i  (arc  AD  +  arc  DC) ; 

i.e.,  Z  ABC  is  meas.  by  i  arc  AC.  q.e.d 

3°.   If  0  lies  without  BA  and  5C,  draw  BOD,  a  diam. 

Since  Z  ABD  is  meas.  by  ^  arc  ^D, 

and  Z  2)5(7  is  meas.  by  i  arc 

Z  ABD  —  Z  DBC  is  meas.  by  i  (arc  ^i)  —  arc  DC) ; 

i.e.,  Z  ABC  is  meas.  by  ^  arc  ^c.  q.e.d 


(162) 

(68) 

(122) 

(Ax.  7) 

(262) 

Q.E.D. 


(1°) 


AD,  I 
DC, I 


(1°) 


PROPORTIONAL  ANGLES.  133 

265.  Definition.   A  segment  of  a  circle  is  the  figure  con- 
tained by  a  chord  and  its  arc. 

266.  Cor.  1.  All  angles  C,D,E,  inscribed      e 
in  the  segment  AEDCB  of  a  circle  are  equal. 

For  each  is  measured  by  half  the  arc       2 
AFB. 

267.  CoR.  2.   An  angle  inscribed  in  a  semicircle  is  a  right 
angle. 

For  it  is  measured  by  |-  a  semicircumference ;  i.e.,  by  \ 
a  circumference,  or  a  quadrant. 

268.  CoR.  3.    The  arc  intercepted  by  an  iyiscribed  angle  is 
double  the  arc  intercepted  by  an  equal  angle  at  the  center. 


Proposition  IV.     Theorem. 

269.  The  angle  formed  hy  a  tangent  and  a  chord 
meeting  at  the  point  of  contact,  is  measured  hy  half 
the  intercepted  arc. 


Given :  An  angle  BAC  formed  by  a  tangent  AB  and  a  chord  AC ; 
To  Prove :  Angle  BAC  is  measured  by  ^  arc  AC. 

Through  C  draw  CD  II  to  AB  (109).     Then 

since  CD  is  II  to  AB,  (Const.) 

3iTG  AD=SiYG  AC,  (195") 

and  Zc  =  Z  J.  (110") 

But  Z  C  is  meas.  by  i  arc  AD,  (264) 

.-.  Z  ^  is  meas.  by  i  arc  AD  ov  ^  arc  AC.        q.e.d. 


134  PLANE   GEOMETRY.— BOOK  IV. 

Proposition  V.     Theorem. 

270.  The  vertical  angles  formed  hy  intersecting 
chords  are  each  measured  hy  half  the  sum  of  the 
intercepted  arcs. 

a 


Given:  Two  chords  AB,  CD,  intersecting  in  E ; 
To  Prove :  Angles  AEC,  bed,  are  each  measured  by  \  (arc 
AC  +  arc  BD). 

Draw  the  chord  BC.     Then 
since  Z  ^  is  meas.  by  i  arc  AC, 

and  Z  (7  is  meas.  by  i  arc  BD,  (264) 

also  Z  AEC  =  Zb  -\-  Zc,  (122) 

Z  AEC  or  Z  BED  is  meas.  by  ^(arc  AC  -{-  arc  BD).    q.e.d. 

Exercise  320.  In  the  diagram  for  Prop.  III.,  if  B  is  an  angle  of  32°, 
how  many  degrees  are  there  in  arc  AC 9 

321.  In  the  diagram  for  Prop.  IV.,  if  CD  is  an  arc  of  102°,  then 
BAC  is  an  angle  of  how  many  degrees  ? 

322.  In  the  diagram  for  Prop.  V,,  if  AEG  is  an  angle  of  25°  and 
J.C  an  arc  of  30°,  how  many  degrees  in  arc  BD  9 

323.  Any  three  points  of  a  circumference  being  given,  how  can 
we  find  other  points  of  it  without  knowing  the  center  ? 

324.  The  opposite  sides  of  an  inscribed  parallelogram  divide  in  the 
same  ratio  the  radii  drawn  perpendicular  to  them. 

325.  If  two  circles  whose  centers  are   0  and  0'  have  a  common 
tangent  AB,  and  OA,  O'B,  be  joined,  these  lines  will  be  parallel. 

326.  If  two  tangents,  PA,  PB,  be  drawn  to  a  circle  whose  center 
is  0,  and  AB,  AO,  be  drawn,  then  will  angle  BAG  =  ^  angle  P. 


PROPORTIONAL  ANGLES.  135 

Proposition  YI.     Theorem. 

271.  The  angle  formed  hy  two  secants  meeting  with- 
out the  circle  is  measured  hy  half  the  difference  of  the 
intercepted  arcs. 


B^ 


E 


)}        ^^^^^ 


Given:  Secants  AB,  AC,  meeting  in  A,  and  intercepting  arcs 
BC,  DE ; 

To  Prove :  Angle  A  is  measured  by  ^(arc  5C  —  arc  DE). 

Draw  the  chord  BD.     Then 
since  Z  52) a  =  Z  ^ -hZ^,  (122) 

/.A  =  /.BDC  —  Z.B.  (Ax.  3) 

But  Z  5I>C7  is  meas.  by  l^arc  J5a, 
and  Z  5  is  meas.  by  \  arc  DE, 
.-.  Z^  is  meas.  by  |-(arc-BC  — arcDjE:).         q.e.d. 

Exercise  327.    In  the  diagram  for  Prop.  VI.,  if  A  is  an  angle  of 
17°  and  DE  an  arc  of  36°,  how  many  degrees  in  sltc  BC  ? 

328.  If  a  polygon  of  an  even  number  of  sides  be  inscribed  in  a 
circle,  the  sums  of  its  alternate  angles  are  equal. 

329.  Find  a  point  equidistant  from  three  given  points.    When  is 
the  problem  impossible  ? 

330.  Under  what  conditions  is  it  possible  to  find  a  point  equidis- 
tant from  four  given  points  ? 

331.  Prove  the  theorem  given  in  Art.  155,  by  means  of  Prop.  III.» 
(171). 

332.  If  an  inscribed  triangle  has  unequal  angles,  the  greater  angle 
intercepts  the  greater  arc. 

333.  If  two  circles  intersect,  their  line  of  centers  produced  will 
bisect  each  of  the  four  arcs. 


136  PLANE   GEOMETRY.  — BOOK   IV. 

Proposition  VII.     Theorem. 

272.  An  angle  formed  by  a  tangent  and  a  secant  is 
measured  by  half  the  difference  of  the  intercepted  arcs. 


Given:  Tangent  AB  and  secant  AC  meeting  in  A  and  inter- 
cepting arcs  BC,  BD ; 

To  Prove :  Angle  A  is  measured  by  i(arc  BC  —  d^xoBD). 

Draw  the  chord  BC.     Then 

since  /.  B  ^  A  A -\- Z.  c,  (122) 

Z  .4  =  Z  5  -  Z  c.  (Ax.  3) 

But  Z  j5  is  meas.  by  ^  arc^C,  (269) 

and  Z  c  is  meas.  by  \  arc  BD,  (264) 

.'.  Z  ^  is  meas.  by  J(arc  BC  —  arc  BD).  q.e.d. 

273.    Cor.    The  angle  formed  by  tivo  tangents  is  measured 
by  half  the  difference  of  the  intercepted  arcs. 


EXERCISES. 

QUESTIONS. 

334.  If  an  angle  A  is  measured  by  two  tliirds  of  a  quadrant,  and  an 
angle  B  =  50°,  what  is  the  ratio  of  Z  ^  to  Z  ^  ? 

335.  By  what  fraction  of  a  quadrant  is  the  vertical  angle  of  an  isos- 
celes triangle  measured,  (1)  if  it  is  twice  as  great  as  a  base  angle  ? 

(2)  if  it  is  I  as  great  ?    (3)  if  it  is  -th  as  great  ? 

n 


EXERCISES.  137 

336.  Two  angles  of  a  triangle  are  measured  by  ^  and  |^  of  a  circum- 
ference, respectively.     How  many  degrees  in  each  of  the  three  angles  ? 

337.  One  angle  of  a  right  triangle  is  measured  by  /^  of  a  circum- 
ference. How  many  degrees  in  the  other  acute  angle,  and  what  is  its 
ratio  to  the  first  ? 

338.  If,  in  the  diagram  for  Cor.  I.  of  Prop.  III.,  ADB  is  an  arc 
of  230°,  how  many  degrees  are  there  in  each  of  the  angles  C,  D, 
and^? 

339.  If,  further,  in  the  same  diagram,  AE  is  an  arc  of  30°,  how 
many  degrees  in  angles  ABE  and  BAE,  respectively  ? 

340.  In  the  diagram  for  Prop.  IV.,  if  BAG  is  an  angle  of  65°, 
how  many  degrees  are  there  in  arc  CD  ? 

341.  In  the  same  diagram,  if  arc  CA  =  ^  arc  CD,  how  many  degrees 
in  Z  BAG? 

342.  In  the  diagram  for  Prop.  V.,  if  AEG  is  an  angle  of  40°, 
how  many  degrees  are  there  in  the  sum  of  the  arcs  AD  and  BC? 

343.  In  the  diagram  for  Prop.  VI.,  if  A  is  an  angle  of  25°,  and 
arc  DE  =  ^  arc  BC,  how  many  degrees  are  there  in  the  sum  of  arcs 
BE  and  CD  ? 

344.  In  the  same  diagram,  if  BD  =  AD,  how  many  degrees  are 
there  in  each  of  the  arcs  BC  and  DE,  if  ZA  =  n  degrees  ? 

345.  In  the  diagram  for  Prop.  VII.,  if  arc  5C=  3  arc  52),  and 
Z.A  —  30°,  how  many  degrees  are  there  in  arc  CD  and  in  Z  JS  ? 

346.  In  the  same  diagram,  if  BA  —  BC,  what  is  the  ratio  of  arc  BG 
to  arc  BD  ? 

347.  If  the  angle  formed  by  two  tangents  is  30°,  how  many  degrees 
are  there  in  each  of  the  intercepted  arcs  ? 

348.  If  a  tangent  be  drawn  at  a  vertex  of  an  inscribed  equilateral 
triangle,  what  angle  will  it  form  with  either  adjacent  side  ? 

349.  If  from  the  same  point  in  a  circumference,  a  side  of  a  square 
and  a  side  of  an  equilateral  triangle  be  inscribed,  the  difference  of 
the  arcs  subtended  by  them  will  be  what  part  of  the  circumference  ? 

350.  If  the  vertical  angle  of  an  inscribed  isosceles  triangle  is  54°, 
what  is  the  ratio  of  the  arc  opposite  that  angle  to  either  of  the  other 
arcs? 

351.  If  the  vertical  angle  were  37°  15'  32",  what  would  those  ratios 
be? 


138 


PLANE   GEOMETRY.  — BOOK  IV. 


PROPORTIONAL  LINES. 

Proposition   VIII.     Theorem. 

274.  A  line  draivn  parallel  to  one  side  of  a  triangle 
and  meeting  the  other  two  sides,  divides  them  pro- 
portionally. 


Given :  DE  parallel  to  5C,  a  side  of  triangle  ABC,  and  meeting 
AB,  AC,m  D,  E,  respectively; 

To  Prove :  DB  :  AD  =  EC:  AE. 

1°.   When  AD  and  DB  are  commensurable. 
Let  BF  \)Q  2i  common  measure  oi  DB  and  AD,  so  that 
BF  can  be  laid  oif  3  times  on  DB  and  7  times  on  AD. 
Through  the  points  of  division  draw  lines  FF^,  etc.,  II  to  BC. 
Since  the  ll's  FF\  etc.,  cut  off  3  and  7  equal  parts  on  DB,  AD, 
respectively,        '  (Const.) 

the  ll's  FF',  etc.,  cut  off  3  and  7  equal  parts  on  EC,  AE, 
respectively ;  (1^1) 

.-.  DB  :AD  =  3:7,  and  EC:  AE  =  3:7,  (225) 

.'.  DB  :  AD  =EC  :AE.         Q.e.d.      (232'") 

2°.  When  AD  and  DB  are  incommensurable. 

Suppose  AD  divided  into  any  number  of  equal  parts  n, 
and  that  DB  contains  this  nth  part  of  AD  m  times, 
with  a  remainder  LB.     Through  L  draw  LL^  II  to  BC. 


PROPORTIONAL  LINES.  139 

Since  AD  and  DL  are  commensurables,        (Const.) 

DJL_m  _ElJ_  .-|^ox 

AD~  n~  Ae' 

DB      m  ,  X        T  EC      m  ,  x' 

.:  —  =  -  +  -,  and  —-  =  -  +  -, 
AD       n      n  AE       n       n 

since  DB  and  J?c  are  slightly  >DL,  EL',  resp. 

X     X 

Now   when  n  is  taken  indefinitely  great,  -,  —  become  in- 
definitely  small, 

.-.  ££  =  E^  Q.E.D.     (255) 

AD       AE 

being  the  limits  of  variables  always  equal. 

275.  CoR.  If  two  sides  of  a  triangle  are  cut  by  a  parallel 
to  the  base,  07ie  side  is  to  either  of  its  parts  as  the  other  is  to 
its  corresponding  part. 

E' T-D 

For  since  DB  :  AB  z=  EC  :  AC,       (274) 
.-.  DB  -\-  AB:  DB,  or  AB, 

=  EC-{-AC:  EC,  or  A C,     (246)  ^ 

i.e.,  AD  :  AB,  or  DB,  =  AE:  AC,  or  EC.    D 

Scholium.  It  is  obvious  that  the  theorems  (274,  275) 
hold  good  when  DE  meets  AB,  AC,  produced  in  either 
direction. 


Exercise  352.   Circumferences  described  on  the  arms  of  an  isosceles 
triangle  as  diameters,  will  intersect  in  the  mid  point  of  the  base. 

353.  Circumferences  described  on  any  two  sides  of  a  triangle  as 
diameters,  will  intersect  in  the  third  side  or  the  third  side  produced. 

354.  If  an  intersecting  circumference  pass  through  the  center  of 
another,  the  angle  in  the  exterior  segment  of  the  latter  is  acute. 

355.  J^JS  is  a  common  exterior  tangent  to  two  circles  touching  at 
P;  draw  PA,  PB ;  APB  is  a  right  angle. 

356.  A  tangent  at  the  mid  point  of  an  arc  is  parallel  to  its  chord. 


140  PLANE   GEOMETRY.  — BOOK  IV. 


Proposition  IX.     Theorem. 

276.  Conversely,  if  a  straight  line  divide  two  sides 
of  a  triangle  proportionally,  that  line  is  parallel 
to  the  third  side. 


F  G 

Given:  In  triangle  ABC,  DE  meeting  AB,  AC,  so  that 
AD  :  DB  =  AE:  EC ; 

To  Prove :  DE  is  parallel  to  BC. 

Produce  AB  to  F  so  that  BF  =  db,  and  through  F  draw 
FG  II  to  DE  to  meet  AC  produced,  in  G. 

Since  FG  is  II  to  DE,  (Const.) 

AD  :  DF  =  AE  :EG.  (274) 

But  AD  :  DB  =  AE  :  EC,  (Hyp.) 

.'.  DB '.DF  =  EC:EG.  (249) 

Now  DB  =  ^  DF,  (Const.) 

.-.  EC  =  ^EG, 

.'.    BCis  II  to  DE.  Q.E.D.       (150) 

277.  Definition.  According  as  a  point  is  taken  in  a  given 
line,  or  in  that  line  produced,  the  distances  of  the  point  from 
the  extremities  of  the  given  line  are  called  internal  or  external 
segments  of  the  line.  Hence  the  given  line  is  the  sum  of 
any  two  internal  segments,  and 

the    difference   of   any  two   ex-    « ' ■ 

ternal  segments. 

Thus  \i  AB  is  divided  in  C  and  produced  to  D,  then  ^5 
is  the  sum  of  the  internal  segments  AC,  CB,  and  the  differ- 
ence of  the  external  segments  AD,  BD. 


PROPORTIONAL  LINES.  141 

Proposition  X.     Theorem. 

278.  The  bisector  of  an  interior  angle  of  a  triangle 
divides  the  opposite  side  into  internal  segments 
having  the  same  ratio  as  the  other  two  sides. 

E 


A..--' 


Given:  In  triangle  ABC,  AD  bisecting  angle  BAG,  and  meeting 
BC'mD; 

To  Prove :  BDiDC  =  AB  :AC. 

Through  C  draw  CE  II  to  AD,  to  meet  ^^  produced,  in  E. 


Since  AD  is  II  to  CE, 

(Const.) 

Z  BAD  =Ze, 

(112") 

and  ZCAD  =Za CE. 

(110") 

But  Z  BAD  =  Z  CAD, 

(Hyp.) 

.'.  Zace  =  Ze, 

(Ax.  1) 

.'.   AE  =  AC. 

(65) 

^ain,  since  AD  is  II  to  CE, 

BD  :DC  =  BA  :  AE. 

(274) 

i.e.,BD  :DC  —  AB  :  AC. 

Q.E.D. 

279.    CoR.    Conversely,  in  A  ABC,  if  D  he  a  point  in  BC 

such  that 

BD  :DC=  AB:AC, 

then  AD  bisects  Z  BAC. 

For  the  bisector  of  that  angle  must  pass  through  D  (278), 
and  must  therefore  coincide  with  AD. 


142  PLANE   GEOMETRY.  — BOOK  IV. 

Proposition  XI.     Theorem. 

280.  The  bisector  of  an  exterior  an£le  of  a  triangle 
divides  the  opposite  side  into  external  segments 
having  the  same  ratio  as  the  other  two  sides. 


Given:  In  triangle  ABC,  AD  bisecting  exterior  angle  CAF 
and  meeting  BC  m  D ; 

To  Prove :  BD  :  DC  =  AB  :  AC. 

Through  C  draw  CE  II  to  AD,  to  meet  BA  in  E. 


Since  AD  is  11  to  CE, 

(Const.) 

Z  FAD  =  Z  AEC, 

(112") 

2ind  Z  CAD  =  Z  ACE. 

(110") 

But  Z  FAD  =  Z  CAD, 

(Hyp.) 

.'.   ZACEz=ZAEC, 

(Ax.  1) 

.'.   AE  =  AC. 

{^5) 

Again,  since  AD  is  II  to  CE, 

BD  :DC  =  BA  :  AE, 

(274) 

i.e.,  BD  :DC  —  AB  :  AC. 

Q.E.D. 

281.    Cor.  Conversely,  in  A  ABC,  if  D  he  a  point  of  BC 
produced,  such  that 

BD  :  DC=  AB  '.AC, 
then  AD  bisects  the  exterior  Z  CAF. 

For  the  bisector  of  that  angle  must  pass  through  D,  and 
therefore  must  coincide  with  AD. 


PROPORTIONAL   LINES.  143 

Proposition  XII.     Theorem. 

\.  If  three  or  more  transversals  passing  through 
the  same  point  mahe  intercepts  upon  two  parallels, 
these  intercepts  are  proportional. 


Given:   Transversals  OA,  OB,  OC,  cutting  the  parallels  AC, 
A'C',  in  A,  B,  c,  and  A',  B\  c',  respectively; 
To  Prove :  AB  :  A'b'  =  BC:  B'c'. 

Through  A',  c',  draw  A'd,  &e,  each  II  to  OB.  Then 

since  A'&  is  II  to  AC,  (Hyp.) 

AO:  A'0=zBO  :B^0  =  CO.C'O.  (274) 

Since  A'n,  c'e,  are  each  II  to  OB,  (Const.) 

DB  =  A^B',  and  BE  =  B'c';  (138) 

also  AB  :  DB  or  A'b'  =  AO  :  A'O,  \ 

[  (274) 

Sind  BC:  BE  OT  B'c' =  C0:  C'O,  ) 

.'.  AB:A'b'  =  BC:B'c'.       q.e.d.      (232'") 

As  this  holds  true  of  the  intercepts  made  by  any  three 
transversals  through  0,  it  holds  true  of  the  intercepts  made 
by  any  number  of  such  transversals. 

283.  Cor.  The  parallel  intercepts  between  any  two  con- 
verging transversals  are  proportional  to  the  intercepts  betweeti 
the  parallels  and  the  common  point. 

YoT  AB  :  A'b' =  A0  :A'0  =  B0  :  B'o.  (Above) 


144 


PLANE   GEOMETRY.  — BOOK  IV. 


SIMILAR  POLYGONS. 

284.  Similar  polygons  are  such  as  are  mutually  equian- 
gular, and  have  the  sides  about  the  equal  angles,  taken  in 
the  same  order,  proportional. 

285.  In  similar  polygons,  similarly  situated  points,  lines, 
or  angles  are  said  to  be  homologous. 


Proposition  XIII.    Theorem. 

286.  Triangles  that  are  inutually  equiangular  are 
similar. 


Given:  In  triangles  ABC,  A'b'c',  angle  yl  equal  to  angle  A', 
angle  B  to  angle  B',  and  angle  C  to  angle  C' ; 

To  Prove :  BC:  B'&  =  AB  :  A'b'  =  AC  :  A^C'. 

Place  A  A'b'&  upon  A  ABC,  so  that  Z  A'  =^  Z  A, 
and  A  a'b'c'  takes  the  position  AB'c'. 

Since  Zb'  =  Zb,  (HyP-) 

B'c'  is  II  to  BC,  (112') 

.-.  BC  :  B'c'  =  AB  :  AB'  =  AC  :  AC' ;  (283) 


i.e.,  BC  :  B'c'  =  AB  :  a'b'  =  AC  :  A'c'. 


Q.E.D. 


287.  Cor.  1.    Two  triangles  are  similar,  if  tivo  angles  of 
the  one  are  respectively  equal  to  two  angles  of  the  other.   (121) 

288.  Cor.  2.    Ttco  right  triangles  are  similar,  if  an  acute 
angle  of  the  one  is  equal  to  an  acute  angle  of  the  other.    (123) 

289.  Scholium.     In  similar  triangles,  homologous  sides 
lie  opposite  equal  angles. 


SIMILAR  POLYGONS.  145 

Proposition  XIV.     Theorem. 

290.  Two  triangles  are  similar,  if  an  angle  of  the  one 
is  equal  to  an  angle  of  the  other,  and  the  sides  about 
these  angles  are  proportional. 


B  C  B'  c' 

Given:  In  triangles  ABC,  A'b'c',  angle  A  equal  to  angle  A^ ; 
also  AB  :A'b'  =  AC:A'c'; 

To  Prove:  Triangle  ABC  i^  similar  to  triangle  A'b'c'. 

Place  A  a'b'c'  upon  A  ABC,  so  that  Z  A'  =^  Z  A, 
and  A  a'b'c'  takes  the  position  AB'c'.     Then 

since  AB  :  A' b' =  AC  :  AC',  (Hyp.) 

B'C'  is  II  to  BC,  (276) 

.-.  Zb  =  Z  b',  and  Zc  =  Z  c'.  (112") 

.-.  A  ^i? C  is  similar  to  A  ^'iJ'c'.    q.e.d.   (286) 

Exercise  357.  In  the  diagram  for  Prop.  XI.,  if  the  bisector  of 
angle  BAC  be  drawn  so  as  to  meet  the  base  in  D',  show  that  BC  is 
divided  into  internal  and  external  segments  having  the  same  ratio. 

Definition.  —  A  straight  line  is  said  to  be  divided  harmonically  if  it  is 
divided  into  internal  and  external  segments  having  the  same  ratio ;  or  if 
it  is  divided  into  three  segments  such  that  the  whole  line  is  to  either  of  its 
outer  segments  as  the  other  outer  segment  is  to  the  inner  segment. 

358.  Show  that  in  the  diagram  for  Exercise  357,  BC  is  divided 
harmonically  according  to  the  first  definition  above,  and  BD^  accord- 
ing to  the  second. 

359.  Show  that  if  D  and  D'  divide  any  line  MN  harmonically 
according  to  the  first  definition,  then  also  M  and  N  divide  DD'  har- 
monically. 

360.  Also  show  that  MD  is  divided  harmonically  according  to  the 
second  definition. 

Geom.  — 10  ' 


146  PLANE   GEOMETRY.  —  BOOK  IV. 


Proposition  XV.     Theorem. 

291.    Triangles  that  have  their  sides  mutually  pro- 
portional are  similar. 


Given:   In  triangles  ABC,  A'b'c',  AB  :  A'b'  =  AC  :  A'c'  = 
BC:  B'c'; 

To  Prove:   Triangle  A' B'c'  is  similar  to  triangle  ABC. 

On  AB,  AC,  take  AD,  AE  =  A'b',  A'c',  resp.,  and  join  DE. 
Since  AB  :  AB  =  AC  :  AE,   (Hyp.  and  Const.) 


DE  is  II  to  BC, 

(276) 

'.  Zd  =  Zb,  and  Ze  =  Zc, 

(112") 

and  AB  :  AD  z=BC:  DE. 

(283) 

But  AB  :  A'b'  =  BC  :  B'c'. 

(Hyp.) 

.'.  AD  :  a'b'  =  DE  :  B'C'. 

(249) 

Now  AD  =  a'b'. 

(Const.) 

.'.   DE  =  B'C'; 

and  AE  =  A'c', 

(Const.) 

.:  AADE  =  AA'B'C', 

(69) 

.-.  Zb'  =  Zd  =  Zb,  and  Z  c'  =  Z  E  =  Z  C,     (Ax.  1) 
.-.  A^'^'C' is  similar  to  A^J?C.      q.e.d.     (286) 

Scholium.  From  this  and  Prop.  XT.  it  is  seen  tliat 
mutually  equiangular  triangles  have  their  homologous  sides 
proportional ;  and  conversely. 


SIMILAR   POLYGONS. 


147 


Proposition  XVI.     Theorem. 

292.  Two  triangles  are  similar,  if  they  have  their 
sides  respectively  parallel  or  perpendicular  to  each 
other. 


Given:   Triangles  ABC,  a'b'c',  having  their  sides  respectively 
parallel  or  respectively  perpendicular  to  each  other ; 

To  Prove:   Triangle  ABC  is  similar  to  triangle  a'b'c'. 

Since  their  sides  are  1|  or  _L  to  each  other,    (Hyp.) 

any  two  homologous  angles  of  these  triangles  must  be  either 
equal  or  supplementary.  (116,  118) 

Hence  there  are  three  conceivable  cases  to  consider. 

1°.   Suppose  all  the  angles  of  the  one  triangle  respectively 
supplemental  to  the  homologous  angles  of  the  other. 

i.e.,  A-\-A'=2i  st.  Z,    B-{-B'=Si  St.  Z,  and  also  C+C' 
=  a  st.  Z. 

2°.   Suppose  two  supplemental  and  one  equal. 

i.e.,  A  =  A',B-{-  b'=  a  St.  Z,  and  C  -\-  C'=  a  st.  Z. 

3°.   Suppose  all  the  angles  are  mutually  equal.* 

i.e.,  A==  A',  B  =b',  and  .:  C  =  c'. 

Each  of  the  first  two  suppositions  must  be  rejected,  since 

*  If  two  are  right  angles,  they  are  equal  and  supplemental. 


148  PLANE   GEOMETRY.  —  BOOK  IV. 

the  sum  of  the  angles  of  two  triangles  cannot  exceed  two 
straight  angles.     Hence  the  third  alone  is  admissible ; 

.-.  A^^C  is  similar  to  Aa'b'c'.       q.e.d.     (287) 

293.    Scholium.   The  homologous  sides  are  those  mutually 
parallel  or  perpendicular. 


Proposition  XVII.     Theorem. 

294.  Two  polygons  are  similar  if  composed  of  the 
same  number  of  triangles  similar  each  to  each  and 
similarly  placed. 


Given:  In  polygons  P  and  P\  triangles  ^JBJi),  ADC,  ACB, 
similar  to  triangles  a'e'd',  a'jd'c',  A'c'b',  respectively,  and  sim- 
ilarly placed; 

To  Prove :  P  is  similar  to  P'. 

1°.   P  and  P'  are  mutually  equiangular. 

For  their  homologous  A  are  either  homologous  A  of 
similar  A,  or  are  like  sums  of  homologous  A  of  simi- 
lar A;  (Hyp.) 

.-.  Ae  —  A  e\  a  EDA  4-  A  ADC  =  A  e'd'a'  -f  A  a'd'c',  etc. 

2°.   The  homologous  sides  of  P  and  P'  are  proportional. 

For  since  the  A  are  similar,  (Hyp.) 

AB  :  A'b'  =  BC  :  B'c'  =  AC  :  A'c'  =  CD  :  C'd',  etc. ; 

.-.  P  is  similar  to  P'.  q.e.d.     (284) 


SIMILAR  POLYGONS, 


149 


Proposition  XVIII.     Theorem. 

295.  Conversely,  two  similar  polygons  may  he  di- 
vided into  the  same  number  of  triangles,  similar  to 
each  other  and  similarly  placed. 


Given:  ABODE,  or  P,  and  A'b'c'd'e',  or  P',  two  similar 
polygons; 

To  Prove:  P  and  P'  may  be  divided  into  the  same  number 
of  similar  triangles  similarly  placed. 

From  A  and  A',  homologous  vertices  of  P  and  P',  draw 
diagonals  AC,  AD,  and  A'c',  A'd',  respectively. 

1°.  The  number  of  triangles  thus  formed  in  P  and  P', 
respectively,  is  the  same.  For  it  is  equal  to  the  num- 
ber of  sides  in  each,  less  two. 

2°.   These  triangles  are  similar  and  similarly  placed. 


For  since  P  is  similar  to  P' 


=  A'P' :  E'd',  I 


(Hyp.) 
(284) 


Ze  =  Z  p', 

and  AE  :ED 

/.  Aaed  is  similar  to  Aa'e'd',  (290) 

and  Z  EDA  =  Z  e'd'a'.  (289) 

But  Z  EDO  =  Ze'd'c',  (Hyp.) 

.-.  Zedc  —  Zeda  =  Ze'd'c'  —  Ze'd'a',  (Ax.  3) 

i.e.,  Zadc  =  Za'd'c'. 

Again,  since  Aaed  is  similar  to  Aa'e'd',  (Above) 

ED:  DA  =2  E'D' :  D'a'.  (285) 

But  ED  :  DC  =  E'D':  D'C',  (Hyp.) 


150  PLANE   GEOMETRY.  — BOOK  IV. 

.'.  DA:DC  =  D'A' :  D'C',  (249) 

.-.  AADCis  similar  to  A  A'D^c'.  (290) 

In  the  same  way,  the  remaining  triangles  of  P  may  be 

proved  similar  to  the  similarly  placed  triangles  in  P'. 

.-.  P  and  P'  may  be  divided  as  stated.         q.e.d. 


Proposition  XIX.     Theorem. 

296.  The  perijneters  of  two  similar  polygons  have 
the  same  ratio  as  any  two  homologous  sides. 


Given:  AB,  A'b',  any  two  homologous  sides  of  two  similar 
polygons,  P,  P',  of  which  j?  andp'  are  the  perimeters ; 
To  Prove :  p  :p' =  AB  :  A'b'. 

Since  P  and  P'  are  similar  polygons,         (Hyp.) 
AB  :  A'b' =  BC:B'C' =  CD  =  C'd',  etc.,  (284) 

.-.   AB  -\-BC-\-CD-\ :  A'B'  +  B'C'  +  C'd'  H =  AB  :  A'b'. 

(251) 

But p  =  AB  +BC -i- CD -\ ,  Sindp'  =  A'B'-{-B'C'+C'D'-\ , 

.-.  p:p' =  AB  :  a'b'.  q.e.d. 

Exercise  361.  Draw  a  diagram  to  show  that  two  figures  may  be 
mutually  equiangular  though  not  similar ;  and  another  to  show  that 
two  figures  may  have  their  sides  mutually  proportional  and  yet  not 
be  similar. 

362.  In  the  diagram  for  Prop,  XVII.,  if  the  numerical  measures  of 
BC  and  B'C  are,  respectively,  16  and  10,  what  is  the  ratio  of  j?  top'  ? 


RATIOS  OF  CERTAIN  LINES.  151 

RATIOS  OF  CERTAIN  LINES. 
Proposition  XX.     Theorem. 

297.  In  a  right  triangle,  if  a  perpendicular  he 
drawn  from  the  vertex  of  the  right  angle  to  the  hy- 
potenuse, 

i°.  The  perpendicular  is  a  mean  proportional 
between  the  segments  of  the  hypotenuse. 

2°.  Each  arm  is  a  mean  proportional  between  the 
hypotenuse  and  the  adjacent  segment. 


Given :  In  a  right  triangle  ABC,  AD  perpendicular  io  BC  from 
the  right  angle  BAC ; 

BD  :DA  =  DA  :  DC. 
BC:AB  =  AB:BD,  and^C:  AC=AC:DC. 


(1° 
To  Prove :    j 


Since  BDA  and  BAC  are  rt.  A,  (Hyp.) 

and  acute  Z  B  is  common  to  both, 
rt.  A  BDA  is  similar  to  rt.  A  BAC.  (288) 

Since  CD  A  and  BAC  are  rt.  A,  (Hyp.) 

and  acute  Z  C  is  common  to  both, 
rt.  A  CD  A  is  similar  to  rt.  A  BAC,  (288) 

.-.  rt.  A  BDA  is  similar  to  rt.  A  CD  A,  (286) 

each  being  similar  to  rt.  A  BAC. 
Since  A  BDA  is  similar  to  A  CD  A, 

BD  :DA=DA:  DC.  Q.E.D.  (284) 

Since  A^^C  is  similar  to  A  BDA  and  CD  A, 
BC:AB=AB:BD,    ] 
mdBC:AC  =  AC:DC.]        ^•^•^-        ^^^^^ 


(297) 


152  PLANE   GEOMETRY.  — BOOK  IV. 

298.  Cor.  If  from  any  point  A  in 
a  circumference,  a  perpendicular  he 
drawn  to  a  diameter  BC,  and  chords 
AB,  AC  J  be  drawn, 

since  BAC  is  a  right  angle,  (267) 

DC:  DA  =z  DA:  DB, 
BC  :  BA  =  BA:  BD,  and  BC:  CA=CA:  CD. 
Hence 

1°.    The  perpendicidar  from   any  point  of  a   circumfer- 
ence to  a  diameter  is  a  mean  proportional  between  the  segments. 
2°.    The  chord  drawn  from  the  point  to  either  extremity  of 
the  diameter  is  a  mean  proportional  between  the  diameter  and 
the  adjacent  segment. 

299.  Definition.  Two  ratios  are  said  to  be  mutually 
inverse  or  reciprocal  when  the  antecedent  and  the  conse- 
quent of  the  one  are,  respectively,  the  consequent  and  the 
antecedent  of  the  other.  Thus  B  :  A  \^  the  inverse  oi  A:  B, 
and  11 :  7  is  the  inverse  of  7  :  11. 

300.  Definition.   If  four  quantities.  A,  B,  C,  D,  are  so 

related  that 

A       D 

A:  C  =  D  :  B,  or  -  =  -, 

'         C      B' 

i.e.,  if  the  first  has  to  the  third  the  inverse  ratio  of  the 
second  to  the  fourth,  the  quantities  are  said  to  be  inversely 
proportional,  while,  as  we  know,  if  the  first  is  to  the  third 
as  the  second  to  the  fourth,  the  quantities  are  directly  pro- 
portional. As  will  be  seen  in  the  next  proposition,  two 
lines  have  their  segments  inversely  proportional  if  a  seg- 
ment of  the  first  is  to  a  segment  of  the  second  as  the  re- 
maining segment  of  the  second  is  to  the  remaining  segment 
of  the  first ;  while  the  segments  would  be  directly  propor- 
tional if  a  segment  of  the  first  were  to  a  segment  of  the 
second  as  the  remaining  segment  of  the  first  to  the  remain- 
ing segment  of  the  second. 


RATIOS   OF  CERTAIN  LINES.  153 

Proposition  XXI.     Theorem. 

301.  If  two  chords  intersect,  their  segments  are  in- 
versely proportional. 


Given :  In  circle  ABB,  chords  AB,  CB,  intersecting  in  0; 
To  Prove :  OA:  OB  =  OC  :  OB. 

Join  AC  and  BB.     Then 

since  Za=Zb,  and  Zb  =  Zc,  (264) 

A  ^  0 C  is  similar  to  Ab  OB,  (286) 

.-.    OA  :  OB  =  OC  :  OB.  Q.E.D.  (289) 

Scholium.  In  deducing  such  proportions  between  pairs  of 
sides  in  similar  triangles,  the  student  may  find  it  useful  to 
remember  that,  as  the  homologous  sides  are  opposite  equal 
angles,  the  greater  side  of  the  one  is  to  the  greater  side  of  the 
other  as  the  lesser  side  of  the  one  is  to  the  lesser  side  of  the 
other. 

Exercise  363.  In  the  diagram  for  Prop.  XI.,  if  a  parallel  to  EC 
cut  AB,  AC,  and  AD,  in  G,  H,  and  L,  respectively,  show  that  AD  is 
divided  by  GL  so  that  AL:AD=GH:BC. 

364.  In  the  diagram  for  Prop.  XII.,  show  that  the  triangles  A' AD 
and  OA'B'  are  similar  to  each  other  and  to  triangle  OAB. 

365.  In  the  same  diagram,  show  that  the  perimeter  of  triangle  OAB 
is  to  that  of  triangle  OA'B'  as  OB  is  to  OB'. 

366.  In  the  diagram  for  Prop.  XX. ,  if  Z  J5  :  Z  (7  =  3  :  5,  how  many 
degrees  in  each  of  the  acute  angles  at  ^  ? 

367.  In  the  same  diagram,  if  the  numerical  measures  of  BC  and 
DC  are  10  and  2,  respectively,  what  is  the  numerical  measure  of  AD  ? 

368.  Find  al§o  the  numerical  measures  oi  AB  and  A  C. 


154  PLANE   GEOMETRY— BOOK  IV. 

Proposition  XXIT.     Theorem. 

302.  If  two  secants  be  drawn  from  a  point  with- 
out a  circle,  these  secants  and  their  external  seg- 
ments are  inversely  proportional. 


Given :  Two  secants  OA,  OD,  cutting  circle  A1)B  in  B,  A,  and 
C,  D,  respectively ; 

To  Prove  :  OA:  OD  =  OC  :  OB. 

Join  AC  and  BD.     Then 

since  /.A  =  /.  D,  and  Z  0  is  common  to  both,  (264) 

A  ^OC  is  similar  to  A  DOB,  (287) 

.-.    OA:OD  =  OC:  OB.  Q.E.D.  (289) 

Exercise  369.  Prove  Prop.  XXI.  by  joining  AD  and  BC^  and 
showing  that  OA:  OD  -  OC:  OB. 

370.  In  the  diagram  for  Prop.  XXII.,  show  that  the  triangles  whose 
vertices  are  at  the  intersection  of  AC  and  BD  are  similar. 

371.  If  two  equal  chords  intersect,  their  segments  are  severally- 
equal. 

372.  If  equal  chords  be  produced  to  meet,  the  secants  thus  formed 
and  their  external  segments  will  be  severally  equal. 

373.  In  the  diagram  for  Prop.  XXI.,  show  that  if  the  chords  are 
equally  distant  from  the  center,  they  are  directly  as  well  as  inversely 
proportional. 

374.  In  the  diagram  for  Prop.  XXII.,  show  that  if  the  secants  are 
equally  distant  from  the  center,  they  are  directly  as  well  as  inversely 
proportional. 

375.  If  two  circles  intersect,  tangents  drawn  to  them  from  any 
point  in  their  common  chord  produced,  will  be  equal. 


EXERCISES.  155 

Proposition  XXIII.     Theorem. 

303.  If  a  secant  and  a  tangent  he  drawn  from 
a  point  without  a  circle,  the  tangent  is  a  mean 
proportional  between  the  secant  and  its  external 
segment. 


Given:    A  tangent  OC  toucliing  circle  ABC  m  C,  and  a  secant 
OA  cutting  ABC  in  B,  A; 

To  Prove :  OA  :  OC  =  OC  :  OB. 

Join  AC  and  BC.     Then 
:•  Za  and  Z.OCB  are  each  meas.  by  |  arc  BC,     (264,  269) 
Z.A=/.  OCB, 
and  Z  0  is  common  to  A  0 AC  and  OBC, 

.-.  AOJLC  is  similar  to  AO^C,  (287) 

.-.    OA:OC=OC:OB.  Q.E.D.      (289) 


EXERCISES. 

QUESTIONS. 


376.  In  the  diagram  for  Prop.  VIIL,  if  DB:  AD  =  S:7,  and 
AC  =  65,  what  is  the  value  of  AE?* 

377.  If  in  the  preceding  question  we  substitute  Vll  for  3,  what  is 
the  value  of  ^^? 

*  Value,  here  and  elsewhere,  stands  for  numerical  measure,  the  unit 
being  left  undetermined.  If  decimals  occur  in  a  result,  it  will  be  sufficient 
to  have  two  places  correct. 


156  PLANE   GEOMETRY.  — BOOK  IV. 

378.  In  the  diagram  for  Prop.  X.,  if  AB,  AC,  BC,  =  9,  7,  12, 
respectively,  what  is  the  value  of  BD  and  of  DC? 

879.  In  the  diagram  for  Prop.  XI.,  if  AB,  AC,  BD,  =  9,  7,  20, 
respectively,  what  is  the  value  ot  BC? 

380.  If  the  angles  at  the  base  of  A  ABC  in  Prop  XI.  are  equal, 
how  is  the  proposition  modified  ? 

381.  If  two  triangles  have  an  angle  of  the  one  equal  to  an  angle 
of  the  other,  and  the  sides  about  another  angle  proportional,  are  they 
necessarily  similar  ? 

382.  In  the  diagram  for  Prop.  XII.,  if  AB,  AC,  A'B,'=  a,  b,  a', 
respectively,  what  is  the  value  of  A'C  ? 

383.  In  the  diagram  for  Prop.  XVII.,  if  AD,  DE,  A  C,  A'D',  =  4, 
3,  5,  3.2,  respectively,  what  are  the  values  of  D'E',  A'C  ? 

384.  In  the  diagram  for  Prop.  XX.,  if  AB,  AC,  BC,  =4,  3,  5, 
respectively,  what  are  the  values  of  AD,  BD,  DC? 

385.  In  the  diagram  for  Prop.  XXI.,  if  ^0  =  f  OD,  and  OB  =  8, 
what  is  the  value  of  OC? 

386.  In  the  diagram  for  Prop.  IX.,  if  BD  is  —  of  AD,  what  part 
is  Ca  of  AG  ?  '"^ 

387.  In  the  diagram  for  Prop.  X.,  if  AB  :  AC,  =  \0:  7,  what  is  the 
ratio  of  ^Z>  to  JS'C  ? 

388.  In  the  same  diagram,  if  AB  =  A  C,  how  many  degrees  are 
there  in  angle  ^C^? 

389.  In  the  diagram  for  Prop.  XI.,  if  EC :  AD,  =2:3,  what  is  the 
ratioof  ^Bto^C? 

390.  In  the  same  diagram,  if  AB  is  equal  to  AC,  where  will  the 
point  E  fall  ? 

391.  In  the  diagram  for  Prop.  XII.,  if  ^C  =  12  and  OB :  OB'  =  8:5, 
what  is  the  value  oi  AD-\-  EC? 

392.  In  the  same  diagram,  if  OB  bisects  angle  AOC,  AB  =  10, 
and  BC=S,  what  is  the  ratio  of  OA'  to  OC  ? 

393.  In  the  diagram  for  Prop.  XIII.,  if  BC  :  B'C>  =  m  :  n,  what  is 
the  ratio  ot  AB  -  AC  to  A'B'  -A'C? 

394.  In  the  diagram  for  Prop.  XVII.,  if  AB  =  18,  and  A'B'  =  12, 
what  is  the  ratio  of  yl  C  to  A'  C  ? 

395.  In  the  diagram  for  Prop.  XX.,  if  BD  :  DA  =  m  :  n,  what  is 
the  ratio  of  the  perimeter  of  triangle  ADB  to  the  perimeter  of  triangle 
ADC? 


CONSTRUCTIONS. 


157 


F  G 


CONSTRUCTIONS. 

304.    To  divide  a  gicen  line  AB  proportionally  to  another 
given    line    CD     divided    in 
E,  F,  G. 

Upon  ad',  making  any 
angle  with  AB,  lay  off  AE', 
e'f',f'g',  g'd',=ce,  ef,  fg, 
GD,  respectively.  Join  BD', 
and  through  e',  f\  g',  draw 
lines  parallel  to  BD',  meeting  AB  in  X,  Y,  Z,  respectively. 
Then  (274), 

AX:  XY:  YZ  :  ZB  =  AE' :  E'F'iF'g':  G'd'=CE  :  EF  :  FG  :  GD. 

If  instead  of  a  divided  line  we  have  numbers  given,  say 
3,  7,  9,  etc.,  we  lay  off  on  AD',  ae'=  3,  e'f'z=  7,  etc.,  and 
AB  will  be  divided  proportionally  to  the  given  numbers. 


305.    To  find  a  fourth  proportional  to  three  given  lines 
A,  B,  C. 


Draw  DE,  DF,  making  any  angle  with  each  other.  Upon 
DE  lay  off  DG,  GE,  equal  to  A  and  B,  respectively,  and  on 
DF  lay  off  DH  =  C.  Join  GH,  and  through  E  draw  EX 
parallel  to  GH,  and  meeting  DF  in  X.     Then 


DG:  GE  =  DH:  IIX,  or  A  :  B  =  C  :  HX. 


(274), 


It  is  obvious  that  if  B  —  C,  we  take   GE  =  DH,  and  we 
obtain  by  this  construction  a  third  proportional  to  A  and  B. 


158 


PLANE    GEOMETRY.  — BOOK  IV. 


/ 

X 

\ 

i 

C                            ED 

A 

B 

306.  To  find  a  mean  proportional  between  two  lines,  A,  B. 
Upon   an  indefinite  line  lay  off 

ED,  CE,  respectively  equal  to  A  and 
B.  Upon  CD  as  diameter  describe 
a  semicircle  DXC;  at  E  draw  EX 
perpendicular  to  CD  to  meet  the 
circumference  in  X.  Then  (312"), 
ED  OV  A:EX  =  EX:  CE  or  B. 

307.  Definition.  A  straight  line  is  said  to  be  divided  in 
extreme  and  mean  ratio,  when  it  is  divided  into  two  seg- 
ments such  that  the  greater  segment  is  a  mean  proportional 
between  the  whole  line  and 

the  lesser  segment.     Thus  if  A 9  b 

AB  is  divided  in  C  so  that 

AB  :  AC  =AC:BC,  then  AB  ...•••■■ ~~~"  •• .. 

is   divided   in  extreme    and  /  \d 

mean  ratio.  /  .-  ■ '    \ 

308.  To  divide  a  line  AB 
in  extreme  and  mean  ratio. 

At  B  draw  5  c  JL  to  AB,  and      ..-•• 

make  5(7=  1^5.  ^ 

From  C  as  center,  with  radius  CB,  describe  a  circumf.  BDE. 

Through  A  and  C  draw  a  secant  meeting  the  circumference 

in  D  and  E. 
From  A  as  center,  with  radius  AE,  describe  an  arc  EX 
meeting  AB  in  X. 

Then  AB  :  AX  =  AX  :  BX.     For 

•••  AB  is  a  tangent  (191),  and  ^z>  a  secant  to  O  £i)^,  (Const.) 

AE  :  AB  =  AB  :  AD,  (303) 

.-.   AE  :  AB  —  AE  =  AB  :  AD  —  AB.  (247) 

But  AB  =  2BC  =  ED,  whence  AD  —  AB  =i  AE  =  AX, 

.-.  AX:BX  =  AB  :  AX,  or  AB  :  AX  =  AX  :  BX.     (307) 


X 


CONSTR  UCTIONS. 


169 


309.    To  divide  a  given  line  AB  harmonically  in  the  ratio 
of  two  lines  C  and  D. 

Upon  the  line  AM, 
making  any  angle  with 
AB,  lay   off   AE=  C ;  f 

and  on  each  side  of  E 
lay  off  EF,  EG,  each    ^ 
equal  to  D.     Join  FB,  GB,  and  draw  EX,  EY,  parallel  to 
GB  and  FB,  respectively.     Then, 


AX  :  BX  =  AE  :EG  =  C  :  D, 
and  AY:BY=  AE  :FE 


(275) 


310.  Upon  a  straight  line  A'b'  to  construct  a  polygon  simi- 
lar to  a  given  polygon. 

Divide  the  given 
polygon  ABODE  into 
triangles  by  diagonals 
drawn  from  vertex  A. 
At  the  extremities  of 
A^B\  make  Zb'  =  Zb,  and  Zb'A'c'  =  Z  BAC.  Then  will 
AA^B'c'he  similar  to  A^^c(288).  In  like  manner  con- 
struct A  A'c'd'  similar  to  Aacd,  and  A'd'e'  similar  to  ADE. 

Polygon  A'b'c'd'e'  is  similar  to  polygon  ABODE.   (294) 

311.  Upon  a  given  line  AB,  to  describe  a  segment  of  a  cir- 
cle such  that  any  angle  inscribed 
in  it  shall  equal  a  given  Z  O. 

At  A,  make  Z  BAD  =  Z  0  (203); 
draw  AG  perpendicular  to  AD  at 
A,  and  draw  EG  perpendicular  to 
AB  Sit  its  mid  point  E.  From  0 
the  intersection  of  AG,  EG,  de- 
scribe arc  ABX.     ABX  is  the  required  segment.  (269) 

Proof.     Z  BAD  =  Z  C,  is  measured  by  ^  arc  ADB,  as  is 
also  any  angle  inscribed  in  BXA. 


■D 


160 


PLANE   GEOMETRY.  — BOOK  IV. 


312.    From  a  given  point  A,  in  or  without  a  given  circum- 
ference BCD,  to  draw  a  tangent  to  BCD. 

Find  0,  the  center  of  BCD,  (173)  and  join  OA.     Then 
1°.    If  A  is  in  the  circumference,  draw  XY  perpendicular 
to  OA  at  A.     XY  is  tangent  to  BCD  at  A.  (191) 


2°.  If  A  is  without  the  circumference,  bisect  OA  in  E. 
From  E  as  center,  with  radius  EA,  describe  a  circumference 
AXY,  cutting  BCD  in  X  and  Y. 

Join  AX,  A  y;  then  AX,  A  Y  are  tangents  to  BCD.   (267, 191) 
Proof.     Join  OX,  0  Y,  and  show  that  OX,  0  Y  are  perpen- 
dicular to  AX,  A  Y,  respectiv^ely. 


EXERCISES. 


THEOREMS. 


396.  The  chords  that  join  the  near  extremities  of  equal  chords  are 
parallel. 

397.  The  opposite  angles  of  a  quadrilateral  inscribed  in  a  circle  are 
supplementary. 

Definition.  Three  or  more  points  are  said  to  be  concyclic  if  a  circum- 
ference can  be  described  through  them.  Thus  the  preceding  theorem  could 
be  enunciated  thus :  A  quadrilateral  whose  vertices  are  concyclic  has  its 
opposite  angles  supplementary. 

398.  If  two  opposite  angles  of  a  quadrilateral  are  supplementary, 
its  vertices  are  concyclic. 

399.  If  AB,  CD,  intersect  in  0,  so  that  AG  .  OC  -  OD  :  OB,  then 
A,  B,  C,  D,  are  concyclic. 

400.  If  OA,  OD,  are  divided  in  B  and  C,  respectively,  so  that 
OA:  0D=  OC:  OB,  then  A,  B,  C,  D,  are  concyclic. 


EXERCISES.  161 

401.  If  an  arc  be  divided  into  three  equal  parts  by  chords  drawn 
from  one  extremity  of  the  arc,  the  middle  chord  bisects  the  angle 
formed  by  the  other  two. 

402.  If,  from  any  point  of  a  circumference,  a  tangent  and  a  chord 
be  drawn,  the  perpendiculars  upon  these  lines  from  the  mid  point  of 
the  intercepted  arc  are  equal. 

403.  The  diagonals  of  a  trapezoid  cut  each  other  in  the  same  ratio. 

404.  If  through  one  of  the  points  of  intersection  of  two  equal  cir- 
cles, any  line  be  drawn  to  meet  the  circumferences,  the  extremities  of 
this  line  are  equally  distant  from  the  other  point  of  intersection. 

405.  If,  in  a  right  triangle,  the  altitude  upon  the  hypotenuse  divides 
it  in  extreme  and  mean  ratio,  the  lesser  arm  is  equal  to  the  farther 
segment. 

406.  In  any  right  triangle,  one  arm  is  to  the  other  as  the  difference 
of  the  hypotenuse  and  the  second  arm  is  to  the  intercept  on  the  first 
arm  between  the  right  vertex  and  the  bisector  of  the  opposite  acute 
angle. 

407.  The  altitudes  of  a  triangle  are  inversely  proportional  to  the 
sides  upon  which  they  are  drawn. 

408.  If  from  an  angle  of  a  parallelogram  ABCD,  a  line  be  drawn 
cutting  a  diagonal  in  E  and  the  sides  in  P  and  Q^  respectively,  then 
will  AE  be  a  mean  proportional  between  PE  and  QE. 

409.  If  from  the  extremities  of  a  diameter,  perpendiculars  be  drawn 
to  any  chord  of  the  circle,  the  feet  of  these  perpendiculars  will  be 
equally  distant  from  the  center. 

410.  Show  that  there  may  be  two,  but  not  more  than  two,  similar 
triangles  in  the  same  segment  of  a  circle. 

411.  If  two  circles  are  tangent  externally,  a  common  exterior  tan- 
gent is  a  mean  proportional  between  their  diameters. 

412.  The  chord  drawn  from  the  vertex  of  an  inscribed  equilateral 
triangle  to  any  point  in  the  opposite  arc  is  equal  to  the  sum  of  the 
chords  drawn  to  that  point  from  the  other  vertices. 

413.  If  two  circles  are  tangent  externally,  lines  drawn  through  the 
point  of  contact  to  the  circumferences  are  divided  proportionally  at 
the  point  of  contact. 

414.  If  a  circle  is  tangent  to  another  internally,  all  chords  of  the 
outer  circle  drawn  from  the  point  of  contact  are  divided  proportionally 
by  the  circumference  of  the  inner  circle. 

Greom.  — 11 


162  PLANE   GEOMETRY.  — BOOK  IV. 

415.  Chords  drawn  from  the  point  of  contact  of  a  tangent  have 
their  segments  made  by  any  chord  parallel  to  the  tangent,  inversely 
proportional. 

416.  If  a  tangent  is  intercepted  between  two  parallel  tangents  to 
the  same  circle,  its  segments  made  by  the  point  of  contact  have  the 
radius  as  mean  proportional. 

417.  If  three  circles  intersect  each  other,  their  three  common  chords 
pass  through  the  same  point. 

418.  If  through  the  mid  point  of  a  side  of  a  triangle  a  line  be  drawn 
intersecting  a  second  side,  the  third  side  produced,  and  a  line  parallel 
to  the  first  through  the  opposite  vertex,  the  line  will  be  divided  har- 
monically. 

PROBLEMS. 

419.  To  a  given  circle  draw  a  tangent  that  shall  be  perpendicular 
to  a  given  line. 

420.  To  a  given  circle  draw  a  tangent  that  shall  be  parallel  to  a 
given  line. 

421.  To  a  given  circle  draw  two  tangents  including  a  given  angle. 

422.  In  a  given  straight  line,  find  a  point  such  that  the  tangents 
drawn  from  it  to  a  given  circle  shall  include  

the  greatest  angle.  '■•••.. 

423.  In  a  chord  produced,  find  a  point  such        /^      "~~~\    ..  ■•' 
that  the  tangent  from  that  point  shall  be  equal     /  ..-A 

to  a  given  line.  ••'■*      \--'y^\ 

424.  From  a  given  center  describe  a  circle     \^^'^^''''^  J^ 
tangent  to  a  given  circle.  ^ -^ 

425.  Describe    a    circumference    passing 

through  a  given  point  and  touching  a  given  circle  in  a  given  point. 

426.  Describe  two  circles  with  given  radii,  so  intersecting  that  their 
common  chord  shall  have  a  given  length  not  greater  than  the  lesser 
diameter. 

427.  With  a  given  radius,  describe  a  circle  tangent  to  two  given 
circles. 

428.  Draw  a  common  exterior  tangent  to  two  given  circles. 

429.  Draw  a  common  interior  tangent  to  two  given  circles. 


EXERCISES.  163 

430.  Find  a  point  such  that  the  tangents  drawn  from  it  to  the  outer 
sides  of  two  tangent  circles  shall  include  a  given  angle. 

431.  Divide  any  side  of  a  triangle  into  two  parts  proportional  to  the 
other  sides. 

432.  Divide  any  side  of  a  triangle  into  three  parts  proportional  to 
the  three  sides. 

433.  From  a  given  line  cut  off  a  part  that  shall  be  a  mean  propor- 
tional between  the  remainder  and  another  given  line. 

434.  Through  a  given  point  within  a  circle,  draw  a  chord  there 
divided  in  the  same  ratio  as  a  given  chord  through  that  point. 

435.  From  a  given  point  without  a  circle  draw  a  secant  divided  by 
the  circumference  in  a  given  ratio. 

436.  From  a  given  point  in  a  given  arc  draw  a  chord  bisected  by  the 
chord  of  the  given  arc. 

437.  In  a  given  circle  place  a  chord  that  shall  be  trisected  by  two 
given  radii  at  right  angles  to  each  other. 

438.  In  a  given  circle  place  a  chord  parallel  to  a  given  chord,  and 
having  to  it  a  given  ratio  not  greater  than  that  of  the  diameter  to  the 
given  chord. 

439.  Through  one  of  the  points  of  intersection  of  two  given  circles 
draw  a  secant  forming  chords  that  are  in  a  given  ratio. 

440.  Inscribe  a  square  in  a  given  triangle. 

441.  Inscribe  a  square  in  a  given  segment  of  a  circle. 

442.  In  a  given  semicircle  inscribe  a  rectangle  similar  to  a  given 
rectangle. 

443.  In  a  given  circle  inscribe  a  triangle  similar  to  a  given  triangle. 

444.  About  a  given  circle  circumscribe  a  triangle  similar  to  a  given 
triangle. 

445.  In  a  given  triangle  construct  a  parallelogram  similar  to  a  given 
parallelogram. 

.  446.    Construct  a  triangle  having  given  the  base,  the  vertical  angle, 
and  the  length  of  the  bisector  of  that  angle. 


164  PLANE   GEOMETRY.  — BOOK  IV. 


LOCI. 

447.  Find  the  locus  of  the  center  of  each  circumference  that  passes 
through  two  given  points. 

448.  Find  the  locus  of  the  center  of  each  circle  that  is  tangent  to  a 
given  circle  at  a  given  point. 

449.  Find  the  locus  of  the  center  of  each  circle  of  given  radius  that 
is  tangent  to  a  given  circle. 

450.  Find  the  locus  of  the  center  of  each  circle  that  is  tangent  to  a 
given  line  at  a  given  point. 

451.  Find  the  locus  of  the  center  of  each  circle  that  is  tangent  to 
t^ro  given  intersecting  lines. 

452.  Find  the  locus  of  the  points  from  which  pairs  of  tangents  of  a 
given  length  may  be  drawn  to  a  given  circle. 

453.  Find  the  locus  of  the  mid  point  of  any  chord  that  passes 
through  a  given  point  in  a  given  circle. 

454.  Find  the  locus  of  the  mid  point  of  any  secant  that  can  be 
drawn  from  a  given  point  to  a  given  circumference. 

455.  Find  the  locus  of  the  vertex  of  any  triangle  constructed  on  a 
given  base,  with  a  given  vertical  angle. 

456.  Find  the  locus  of  a  point  whose  distances  from  two  given 
points  are  in  a  given  ratio. 

457.  Find  the  locus  of  a  point  whose  distances  from  two  given 
straight  lines  are  in  a  given  ratio. 

458.  Find  the  locus  of  a  point  the  sum  of  whose  distances  from  two 
given  straight  lines  is  equal  to  a  given  line. 

459.  Find  the  locus  of  a  point  the  difference  of  whose  distances 
from  two  given  straight  lines  is  equal  to  a  given  line. 

460.  Find  the  locus  of  the  points  that  divide  the  chords  of  a  given 
circle  so  that  the  rectangle  of  their  segments  is  equal  to  a  given  square. 


Book  Y. 
areas  and  their  comparison. 


5i*i< 


QUADRILATERALS. 

313.  The  area  of  a  plane  figure  is  the  quantity  of  its 
surface  as  measured  by  the  unit  of  surface,  or  is  the  numer- 
ical measure  of  that  quantity. 

314.  Figures  that  are  not  similar  but  have  equal  areas 
are  said  to  be  equivalent. 

315.  The  base  of  a  polygon  is  any  side  on  which  we 
choose  to  regard  it  as  constructed. 

316.  The  altitude  of  a  polygon  is  the  perpendicular  dis- 
tance to  the  base  from  the  remotest  vertex  or  from  a  side 
parallel  to  the  base  or  the  base  produced. 


B      D      A   P   B 


A      P 


B      JP 


Thus  in  each  of  the  figures  above,  the  perpendicular 
CP  is  the  altitude  of  the  figure  when  ^^  is  taken  as  base. 
It  is  obvious  that  two  triangles  having  their  bases  in  the 
same  line  and  the  opposite  vertex  common,  as  ACB  and  ACD, 
have  the  same  altitude. 

165 


166 


PLANE   GEOMETRY.  — BOOK    V. 


Proposition  I.     Theorem. 

317.   Rectangles   with   equal  altitudes  are  to  each 
other  as  their  bases. 


B       A' 


B  E 


D' 


B' 


C 


Given :  Two  rectangles  AC,  A'c',  with  equal  altitudes  AB,  a'b'j 
and  bases  BC,  b'c' ; 

To  Prove :  Kectangle  AC  :  rectangle  A'c'=bc  :  B'c'. 

1°.   When  BC  and  B'c'  are  commensurable. 

Let  BE  be  a  common  measure  oi  BC  and  B'c',  so  that  BE 
can  be  laid  off  5  times  on  B'c'  and  8  times  on  BC. 

Through  each  point  of  division  draw  perpendiculars  to 
the  opposite  side  of  the  rectangle.  The  figures  thus  formed 
are  equal  rectangles.  (144) 

Since  BC  and  B'c'  contain  8  and  5  parts,  respectively,  each 
equal  to  BE,  (Const.) 

AC  and  A'c',  resp.,  contain  8  and  5  parts,  each  equal  to  AE. 

.-.  BC  :  B'C'=  8  :  5,  and  AC  :  A'c'=  8:5;        (225) 

.-.  rect.  ^C:  rect.  ^'6''=i?C':  j5'C''.    q.e.d.   (232'") 

2°.   When  BC  and  B'c'  are  incommensurable. 

Suppose  b'c'  divided  into  any  number  of  equal  parts  n, 
and  that  BC  contains  this  nth  part  of  B'c'  m  times  with 
a  remainder  EC.     Draw  EF  perpendicular  to  BC. 

Since  BE  and  B'c'  are  commensurable,      (Const.) 


BE 

Wc' 


m 
n 


rect.  AE 
rect.  A'C'^ 


(1°) 


QUADRILATERALS.  167 

BC      m      X        ,    rect.  AC        ma?' 

.*•  —7—;  =  — h-)  and    — 7—,  —  — | — , 

B^G'      n      n  rect.  A'C'      n       n 

since  BC  and  ^C  are  slightly  >  BE  and  AE,  respectively. 


Now,  when  n  is  taken  indefinitely  great,  -  and  —  become 
dl; 
rect.  AC       BC 


indefinitely  small ; 


Q.E.D.       (236) 


rect.  ^'C'     ^'C' 
being  the  limits  of  variables  always  equal. 

318.  CoR.  1.  Rectangles  ivith  equal  bases  are  to  each  other 
as  their  altitudes. 

For  (317)  the  equal  bases  may  be  taken  as  altitudes,  and 
the  altitudes  as  bases. 

319.  Scholium.  Since  a  rectangle  is  determined  by  its 
base  and  altitude  (144),  that  is,  by  any  two  adjacent  sides, 
as  AB,  BC,  we  employ  the  expression,  the  rectangle  contained 
by  AB  and  BC,  or  more  briefly,  the  rectangle  AB,  BC,  to  de- 
note the  rectangle  determined  by  AB  and  BC ;  and  use  the 
symbol  rect.  AB  •  AC,  or  simply  AB  •  BC,  for  either  of  these 
expressions.  Since  a  square,  again,  is  determined  by  its 
base,  i.e.,  by  a  side,  Ave  employ  the  expression,  the  square 
of  AB,  or  the  symbol  AB  ,  an  abbreviation  for  AB  •  AB,  to 
denote  the  square  whose  base  is  AB. 

320.  Definition.  The  symbol  =c=,  to  be  read,  is  equiva- 
lent to,  is  the  symbol  of  equivalence. 

321.  Cor.  2.  If  four  lines,  A,  B,  C,  D,  are  in  proportion, 
the  rectangle  contained  by  the  extremes  is  equivalent  to  that 
contained  by  the  means. 


B.n  =  A:B,  I 

B'D  =  C  :D  =  A:B,)      ^        ^ 


For  since  rect.  A-D  :  rect.  B  -D  =  A:B, 
and  rect.  B  •  C  :  rect. 
rect.  A'B  :  rect.  B  •  C ^veat.  B-D  :  rect.  B  •  D,   (232'") 
.-.  rect.  A  -D  =0=  rect.  B  •  (7. 


168 


PLANE   GEOMETRY.  — BOOK    V. 


322.  Cor.  3.     If  A,  B,  c,  are  lines  such  that 

A:  B  =  B  :C, 
then  B^  =  rect.  A  -  C.  (321) 

That  is,  the  square  of  a  mean  proportional  between  two  lines 
is  equivalent  to  the  rectangle  contained  by  those  lines. 

323.  The  U7iit  of  area  is  the  square  having  as  base  the 
linear  unit.     Thus  if  the  base  AB  of  the  r. 
square  ^C  is  equal  to  the  linear  unit,  then 
the  square  AC  is  the  unit  with  which  all 
areas  are  compared. 

324.  As  already  defined,  the  numerical 
measure  of  a  quantity  is  the  number  that  ^  ^ 
shows  how  many  times  the  quantity  contains  its  unit ;  in 
other  words,  it  is  the  ratio  of  the  quantity  to  its  unit.  As 
regards  triangles,  it  is  customary  to  denote  the  numerical 
measure  of  a  side  by  means  of  the  small 
letter  corresponding  to  the  capital  designat- 
ing the  opposite  angle.  Thus  in  A  ABC, 
we  employ  a,  b,  c,  to  denote  the  numerical 
measures  of  BC,  AC,  AB,  respectively.  As  regards  polygons 
of  more  sides  than  three,  there  is  no  such  convention,  but 
we  specify  AB  =  a,  BC  =  b,  CD  =  c,  and  so  on.  Wherever  it 
may  occur,  henceforth,  such  an  expression  as  the  product 
of  A  and  B  is  to  be  understood  as  a  con- 
venient abbreviation  for  the  product  of 
the  numerical  value  of  A  by  that  of  B. 
Great  care  should  be  taken,  however,  not 
to  forget  the  real  meaning  of  such  abbre- 
viations. Beginners  are  often  confused 
by  the  careless  use  of  such  expressions  as,  leiigth  multiplied 
by  breadth  gives  area,  forgetting  that  what  is  meant  is: 
the  numerical  measure  of  the  length  multijoled  by  that  of  the 
breadth  gives  as  result  the  numerical  measure  of  the  area,  as 
we  find  explained  in  Art.  326. 


QUADRILA  TERALS. 


169 


Proposition  II.     Theorem. 

325.  Rectangles  are  to  each  other  as  the  products  of 
the  numerical  measures  of  their  altitudes  and  bases. 


II 


c' 


Given:  Two  rectangles  AC,  A'c',  with  altitudes  AB,  A'b', 
and  bases  BC,  B^c',  respectively,  whose  numerical  measures  are, 
respectively,  a,  a',  and  b,  b' ; 

To  Prove :  Eectangle  AC  :  rectangle  A'c'  =  a  xb:a'  x  b'. 

Construct  a  rectangle  EG  with  altitude  EF  =  A'b',  and 

base  FG  =  BC,  and  let  the  numerical  measures  of 

AC,  A'c',  EG,  be  X,  y,  z,  respectively. 

Since  AC  and  EG  are  rectangles  with  equal  bases,  (Const.) 

AC:EG  =  AB  :  EF,  OT  X  :  z  =  a  :  a' ;       (318,  232') 

'.'EG  and  a'c'  are  rectangles  with  equal  altitudes,  (Const.) 

EG  :  A'C'  =  FG  :  b'c',  ovz:y=b:b'.     (317,  232') 

From  these  numerical  proportions  we  obtain     (242) 

x:y=  a  xb:a'  X  b', 

.'.  rect.  AC  :  rect.  A'c'  =  axb  ;  ft'  x  b'     q.e.d.      (232") 

326.  Cor.  The  area  of  a  rectangle  is  measured  by  the 
product  of  its  base  and  altitude. 

For  if  ^  Cbe  any  rectangle, 
and  S  the  unit-square,  then 
•.•  area^C:S  =  a  x  6:1  xl, 

area  ^C=  S  xab; 
i.e.,  the   area  of  AG  is  ab 

times  the  unit-square  S.     If,  for  example,  the  numerical 
measures  of  AB  and  BC  are  5  and  7,  respectively,  then  the 


170  PLANE   GEOMETRY.  — BOOK    V. 

numerical  measure  of  the  area  of  ^C  is  35 ;  that  is,  the  area 
of  AC  is  equal  to  35  unit-squares. 

In  the  enunciation  of  this  corollary,  as  elsewhere,  the 
term  area  is  for  brevity  used  for  numerical  measure  of  the 
area;  that  is,  the  number  of  unit-squares  to  which  the  sur- 
face in  question  is  equivalent. 


Proposition  III.     Theorem. 

327.  Any  parallelogram  is  equivalent  to  the  rect- 
angle having  the  same  base  and  altitude. 

A         E  D         F 


B  C 

Given :  A  parallelogram  AC  and  a  rectangle  EC,  with  the  same 
base  and  altitude  BC,  EB  ; 

To  Prove:  Parallelogram  AC  \^  equivalent  to  rectangle  EC. 

Since  AC  and  EC  are  parallelograms,         (Hyp.) 
AB  z=  DC,  and  BE  =  CF,  (1^6) 

.-.   Tt./\ABE  =  itADCF.  (72) 

But  ABCF  —  A  ^i?^^rect.  EC, 
and  ABCF  —  AZ)C'i"<>par'm  AC, 

.'.  par'm  JC=o=rect. -EC.      q.e.d.      (Ax.  3) 

328.  Cor.  1.  Parallelograms  with  equal  bases  and  equal 
altitudes  are  equivalent.  For  each  is  equivalent  to  the  same 
rectangle.  (327) 

329.  Cor.  2.  Parallelograms  icith  equal  altitudes  are  to 
each  other  as  their  bases;  and  those  ivith  equal  bases  to  each 
other  as  their  altitudes.  For  they  are  as  the  rectangles  hav- 
ing those  bases  or  altitudes. 

330.  Cor.  3.  The  area  of  any  parallelogram  is  equal  to 
the  product  of  its  base  and  altitude. 


QUADRILATERALS. 
Proposition  IV.     Theorem. 


171 


331.  Any   triangle   is   equivalent  to   orie   half  the 
rectangle  contained  by  its  base  and  altitude. 


Given  :  A  triangle  ABC,  having  a  base  BC  and  altitude  AD  ; 
To  Prove :  Triangle  ABC  is  equivalent  to  i  rect.  AD  -  BC. 

Complete  the  parallelogram  ABCE. 
Then  since  ^C  is  a  diagonal  of  par'm  BE, 

Aabc=Aace,  (140) 

.-.  A  ^^C  =0=1  par'm  ^^, 

Q.E.D.      (327) 


.-.  A  ABC  ^  i  rect.  AD  -  BC. 


332.  Cor.  1.  Triangles  with  equal  bases  and  equal  alti- 
tudes are  equivalent. 

333.  Cor.  2.  Triangles  with  equal  altitudes  are  to  each 
other  as  their  bases;  and  those  with  equal  bases,  as  their 
altitudes. 

334.  Cor.  3.  Triangles  are  to  each  other  as  the  products 
of  their  bases  and  altitudes. 


Exercise  461.  Prove  Prop.  III.,  when 
the  upper  base  of  the  rectangle  lies  with- 
out that  of  the  given  parallelogram,  as 
in  the  accompanying  diagram. 

462.  Prove  the  same  proposition  when 
the  upper  and  lower  bases  lie  without 
each  other,  though  in  the  same  lines. 


172  PLANE   GEOMETRY.— BOOK    V. 

Proposition  V.     Theorem. 

335.  The  rectangle  contained  hy  a  line  and  the 
sum  or  difference  of  other  two  lines,  is  equivalent 
to  the  sum  or  difference  of  the  several  rectangles 
contained  hy  that  line  and  the  other  two  lines. 


Given :  A  line  AB  and  another  line  BD  equal  %o  BC  ±  CD  ; 
To  Prove :  rect.  AB  -  BD  =c=  rect.  AB'  BC  ±  rect.  AB  •  CD. 
Suppose  the  rectangles   AB  •  BDJ  etc.,  duly  constructed. 

(206) 
1°.    Let  BD  =  BC-\-  CD.     Then,  in  left-hand  Fig., 

rect.  AD  =  rect.  AC  -\-  rect.  ED,  (Ax.  9) 

i.e.,  rect.  AB  •  BD  =  rect.  AB  -  BC  +  rect.  EC  •  CD,     (319) 

.-.  rect.  AB'  BDo=  rect.  AB  -  BC  -\-  rect.  AB  -  CD ,     q.e.d. 

(since  EC  =  AB).  (136) 

2°.   Let  BD  =  BC  —  CD.     Then,  in  right-hand  Fig., 

since  rect.  AB ..  BD  -\-  rect.  AB  -  CD<^ rect.  AB  -  BC,     (1°) 

rect.  AB  •  BD  ^rect  AB  •  BC  —  rect.  AB  •  CD.   q.e.d.   (Ax.  3) 

Scholium.  Let  a,  b,  c,  denote  the  numerical  measure  of 
AB,  BC,  CD,  respectively ;  then  by  substitution  we  obtain 
the  well-known  algebraic  formula 

a{b  ±c)  =  ab  ±  ac. 

336.  Cor.  1.  The  square  of  the  sum  or  difference  of  two 
lines  is  equivalerit  to  the  sum  of  the  squares  of  those  lines  plus 
or  minus  twice  their  rectwigle. 


QUADRILATERALS. 


173 


(319) 
(335) 


thus 


Let  A  and  B  be  the  lines.     Then 

{A±By^  rect.  {A±B)'{A±  B), 

=0=  rect.  A(A  ±  B)±  rect.  B(A  ±B) 
=(^A^±  rect.  ^  .  ^  ±  rect.  B  -  A  +  B'^ 
^A^-{-B^±2TeGt.A'B, 

Scholium.   This  result  may  be  expressed  algebraically 
{a±by=a'±2ab-\-b\ 

337.  Cor.  2.  The  rectangle  contained  by  the  sum  and  dif- 
ference of  two  lines  is  equivalent  to  the  difference  of  the  squares 
of  the  lines. 

Let  A  and  B  be  the  lines.     Then 

{A-{-B)'{A—  B)^  A{A  —  B)-\-B{A  —  B), 
^  A^  —  A-B  +  B'A—  B% 
=^A'-B\ 


}  (335) 


N 


Scholium.   This  result  may  be  expressed  algebraically 
thus :  (a  +  6)  (a  -  6)  =  a^-  b\ 

The  above  corollaries,  pure  deductions  from  Prop.  V.,  be- 
come obvious  to  inspection  in  the  accompanying  diagrams. 

In  the  first,  we  see  that  the  square  of 
MP,  the  sum  of  MN  and  NP,  is  made  up 
of  the  squares  Q  and  R  and  the  two 
equal  rectangles  s,  s. 

In  the  second,  we  see  that  the  square 
of  MP,  the  difference  of  MN  and  NP, 
is  less  than  NMN't,  the  sum  of  the 
squares  of  MN  and  NP,  by  the  figure 
NSN',  which  =  2  P  r  =  2  rect.  MN  •  NP. 

In  the  third,  we  see  that  if  from  the 
square  of  MP  we  take  the  square  of  MN, 
we  obtain  the  figure  QTP,  which  is  equiva- 
lent to  QS  or  {MP  +  PN)  •  (MP  —  PN), 
where  PN  =  MP  —  MN. 


N' 


M 
P' 


S 


M 


N 


T 


M     N 


174 


PLANE   GEOMETRY.  — BOOK   V. 


Proposition  VI.     Theorem. 

338.  A  trapezoid  is  equivalent  to  the  rectangle 
contained  by  its  altitude  and  half  the  sum  of  its 
parallel  sides. 


Given :  A  trapezoid  ABCD,  with  bases  AB,  CD,  and  altitude  CE  ; 
To  Prove :  ABCD  is  equivalent  to  rect.  CE  •  ^{AB  +  CD). 


Join  AC. 

Since  ABCD  o^  A  ABC  -\- A  ADC, 
hut  A  ABC  <>=i  rect.  AB  •  CE,  ^ 
and  AADC^^  rect.  CD  -  CE,  > 

ABCD^lQdt.  CE-i{AB  +  CD). 


(Ax.  9) 
(331) 

Q.E.D.      (335) 


Scholium  1.    The  above  proposition  may  be  expressed 
under  the  form : 

The  area  of  a  trapezoid  is  measured  by  the  product  of  its 
altitude  by  half  the  sum  of  its  bases. 


339.  Scholium  2.  The  area  of  any  polygon  may  be  found 
by  dividing  it  into  triangles,  and  find- 
ing the  areas  of  tliese  (331).  But  the 
method  generally  employed  is  to  draw 
the  most  convenient  diagonal  of  the 
figure,  and  draw  to  it  perpendiculars 
from  the  other  angular  points.  The  figure  is  thus  divided 
into  right  triangles,  rectangles,  or  trapezoids,  the  areas  of 
which  are  easily  found. 


Q  UA  DRILA  TERALS.  175 

Proposition  YII.     Theorem. 

340.  Any  circuinscribed  polygon  is  equivalent  to 
the  rectangle  contained  hy  half  its  perimeter  and 
the  radius  of  the  inscribed  circle. 


Given:  A  polygon  ABODE  circumscribed  about  a  circle  with 
radius  OP  ; 

To  Prove:  ABODE ^iect.OP'i(AB+BC-{-CD-{-DE-{-EA). 

Join  OA,  OB,  etc.,  and  draw  OP^  1.  to  BO,  OP^  ±  to  OD,  etc. 

Since  AAOB=o^ rect.  OP  •  AB,  >> 

and  Aboo  =0=^  rect.  OP2  -DO,  i  (331) 

and  similarly  of  the  other  triangles,  J 

ABODES  rect.  OP  '  ^(AB  +  BO+OD-\-DE-{-EA),    Q.E.D. 

since  ABODE  =  the  sum  of  all  the  A,       (Ax.  9) 

and  OP  =  OP2  =  OP3,  etc.  (162) 

Scholium.  In  the  case  of  the  triangle,  the  most  frequent 
application  of  this  theorem  may  be  stated  as  follows : 

The  area  of  a  triangle  is  equal  to  the  product  of  half  its  per- 
imeter and  the  radius  of  the  inscribed  circle. 

Exercise  463.     lu  diagram  for  Prop.  VI. ,  show  that 
rect.  AB  -  (AE  -  EB)  ^  AE^  -  EB^. 

464.  In  the  same  diagram,  show  that  rect.^^-(^B+  EBy>SB^ —lElB^. 

465.  In  the  same  diagram,  if  Z  ^  =  62°  54'  23",  what  must  Z  Z>  be 
in  order  that  A^  B,  C,  and  D  may  be  concyclic  points  ? 


176 


PLANE   GEOMETRY.  —  BOOK   V. 


Proposition  VIII.     Theorem. 

341.  Triangles  that  have  an  angle  of  the  one  equal  to 
an  angle  of  the  other,  are  to  each  other  as  the  rectangles 
contained  by  the  sides  including  those  angles. 


Given:  Two  triangles,  ABC,  A'b'&,  having  angle  A  equal 
to  angle  A' ; 

To  Prove:  Triangle  ^£C  :  triangle  A'b'c' =:  lect.  AB - 
ACiiect.  A'B'  •  A'c'. 


t.  BD  •  AC  =  AB  :  BD,    » 

[    (317) 
^'D'  'A'&  =  A'B':B'D',) 


Draw  BD,  B'd',  ±  to  AC,  A'c',  resp. 

Since  Z  A  =  Z  A'  (Hyp.),  and  A  D,  D',  are  rt.  A, 

A  BDA  is  similar  to  Ab'd'a' ;  (286) 

.-.   AB  :  BD  =  A'B'  :  B'd'. 

Since  rect.  AB  -AC:  rect.  BD  ■  AC  =  AB  :  BD, 

and  rect.  A'b'  -  A'c' :  rect.  B' 

Tect.AB  'AC:A'B''A'c'=vectBD'AC:TeGt.B'D'-A'c'.  (232'") 

BntAABC  :AA'B'c'=Tect.  BD  -  AC  :  rect.  B'd'  -  A'c';  (334) 

.-.  A  ABC:  A  A'b'C'  =  rect.  AB  -  AC  :  rect.  A'b'  -  A'c'. 

Q.E.D.      (232"') 

Scholium.     The  theorem  may  be  expressed  also  under 

the  form  :   The  areas  of  triangles  that  have  an  angle  of  the 

one  equal  to  an  angle  of  the  other,  are  as  the  products  of  the 
sides  about  those  angles. 


QUADRILATERALS.  177 

Proposition  IX.     Theorem. 

342.  Similar    triangles  are  to  each  other  as  the 
squares  of  their  homologous  sides. 


Given:   Similar  triangles   ABC,   A'b'c',  having   AB  :  AC  = 
A^B'-.A'C'; 

To  Prove :  Triangle  ABC  :  triangle  A'b'c'  =  AB^iHF^  etc. 
Since  Ib^  :  rect.  AB  -  AC=  AB  :  AC, 


and  A^'  :  rect 


Ct.  AB  '  AV=  AB  :  AC,        ") 

.  A'B'  'A'C'  =  A'B':A'C')  ^ 


AB  :  TW  =  rect.  AB  -  AC  :  rect.  A^B^  -  A'c',   (249,  244) 

{since  AB  :  AC  =  A' B' :  A' C'.)  (Hyp.) 

But  A  ABC :  A  A'b'c'  =  Yect  AB  '  AC  iTect.  A'B' '  A'c',   (341) 

(since  Z  ^  =  Z  ^';) 

.-.  AABCiAA'B'c'  =  AB^:A^\        Q.E.D.      (232'") 

343.  Cor.  Similar  triangles  are  to  each  other  as  the 
squares  of  any  homologous  lines. 

Exercise  466.  Show  that  the  square  of  the  sum  of  two  lines 
diminished  by  the  square  of  their  difference  is  equivalent  to  four 
times  the  rectangle  contained  by  the  two  lines. 

467.  In  the  diagram  for  Prop.  IX.,  if  CD,  CD',  be  drawn,  making 
the  same  angle  with  AB,  A'B',  resp., /\  ABC:  A  A'B'C  =  VD'^  :  C'D''^' 

468.  In  the  same  diagram,  what  should  be  the  ratio  of  AB  to  A'B' 
so  that  A  ABC  may  have  twice  the  area  of  A  A'B'C  F 

Geom.  — 12 


178  PLANE   GEOMETRY.— BOOK    V. 

Proposition  X.     Theorem. 

344.  Any  two  similar  polygons  are  to  each  other  as 
the  squares  of  their  homologous  sides. 

D 

D' 


A  B  A'  B' 

Given:  Similar  polygons  ABODE  or  P,  and  A'b'c'd'e'  or  P', 
and  AB  homologous  to  a'b' ; 

To  Prove :  P  :  P' =  AB^ :  A^\ 

Draw  diagonals  from  A  and  A',  dividing  the  polygons  into 
the  same  number  of  homologous  triangles.  (295) 

Since  any  homologous  pair  of  these  triangles,  as  ABC, 
A'b'c',  or  ACD,  A'c'd',  are  as  the  squares  of  any 
homologous  sides,  (342) 

.  AABC:Aa'b'&  =  AACD  :Aa'&D'  =  AB^:  A^^,  etc.; 

.-.  AABC-\-AACD-^AADE:AA'B'C'-\-  AA'C'd' 

+  AA'd'e'  =  Ab':A^';  (251) 

.-.  P  :  P'  ~  AB~ :  J^l  Q.E.D. 

345.  Cor.  1.  Similar  polygons  are  to  each  other  as  the 
squares  of  any  homologous  lines. 

346.  Cor.  2.  The  homologous  sides  of  any  two  similar  2wly- 
gons  are  as  the  square  roots  of  the  areas  of  those  polygons. 

Exercise  469.  In  the  diagram  for  Prop.  X.,  if  AB :  A'B'  =  m:  7i, 
what  will  be  the  ratio  of  the  square  described  on  AD  to  that  described 
on  A'D'  ? 

470.  In  the  same  diagram,  if  P:  P'=m:  n,  what  is  the  ratio  of 
any  two  homologous  lines  in  the  figures,  as  ^O  and  A'C'f 


QUADRILATERALS. 


179 


Proposition  XI.     Theorem. 

347.  In  a  right  triangle,  the  square  of  the  hypot- 
enuse is  equivalent  to  the  sum  of  the  squares  of 
the  arms.  _ 


Given:  A  right  triangle  ABC,  having  the  right  angle  at  A  ; 
To  Prove :     BG^  is  equivalent  to  Zb  -\-  AC  . 

Draw^D±to^C. 


since  BC:AB  —  AB:  BD, 

rect.  BC 'BD=^AB^ ; 
^mCQ  BC :  AC  =  AC '.  DC, 
rect.  BC  '  DC=o=AC^,' 
rect.  BC'BD  -\-  rect.  BC »  DC<^~AB^  +  7c' 
i.e.,  TeotBC'(BD+DC) 


AB^  -\-Ac\' 


.e.,  BG 


AB' 


-\-Ac\ 


(297) 
(321) 
(297) 
(321) 
(Ax.  2) 
(335) 

Q.E.D. 


348.  Cor.  If  any  similar  polygons  are  constructed  on  the 
sides  of  a  right  triangle,  that  on  the  hypotenuse  is  equivalent 
to  the  sum  of  those  on  the  arms. 

349.  Definitions.  The  projection  of  a  point  upon  an 
indefinite  straight  line  is  the  foot  of  the  perpendicular 
drawn  from  the  point  to  the  line. 

Thus  P  is  the  projection  of  the  point  A  upon  MN. 


180 


PLANE   GEOMETRY.  — BOOK    V. 


TlnQ  projection  of  a  finite  straight  line  upon  an  indefinite 
straight  line  is  the  intercept  between  the  projections  of  the 
extremities  of  the  line. 

Thus  PQ  is  the  projection  of  AB  upon  MN.  If  B,  one  of 
the  extremities  of  AB,  is  in  MN-,  then  B  coincides  with  its 
own  projection,  and  PB  is  the  projection  of  AB. 


Propositiox  XII.     Theorem. 

350.  In  any  triangle,  the  square  of  a  side  subtend- 
ing an  acute  angle  is  less  than  the  sum  of  the 
squares  of  the  other  sides  by  twice  the  rectangle  con- 
tained hy  either  of  these  sides  and  the  projection 
upon  it  of  the  other  side. 


Given:  c,  an  acute  angle  of  triangle  ABC,  and  PC,  the  pro- 
jection on  BC  oi  AC; 

To  Prove :  AB^  is  equivalent  to  bO^  +  Ic^  —  2  rect.  BC  -  PC. 

According  as  P  is  in  J5C  or  in  ^C  produced, 
PB  =  BC  —  PC,  or  —PC  —  BC.     In  either  case, 

PB^^BC^-\-P(f-2  rect.  BC'PC;  (336) 

.-.  PB^  4-  AP^  =o  PC^  +  PC^  +  AP^  —  2  rect.  BC  -  PC.    (Ax.  2) 

But  PB^  +  AP^^  ab\  and  PG^  +  Jp^  =o  AC';    (347) 

.-.  Zb^  =o=PC/^  + Ic^  — 2  rect.  PC -PC.  q.e.d. 


QUADRILATERALS.  181 

Proposition  XIII.     Theorem. 

351.  In  an  obtuse  triangle,  the  square  of  the  side 
subtending  the  obtuse  angle  is  greater  than  the  sum 
of  the  squares  of  the  other  sides  by  twice  the  rect- 
angle of  either  of  these  sides  and  the  projection  upon 
it  of  the  other  side. 


Given :  c,  the  obtuse  angle  of  triangle  ABC,  and  PC,  the  projec- 
tion on  BC  of  AC; 

To  Prove :  AB^  is  equivalent  to  'bc^  +  AC^  +  2  rect.  BC  •  PC. 

Since  PB  =  BC  -{-  PC, 

PB^^'BC^-\-PC^-h2Tect  BC'PC;  (336) 

.-.  PB^  +  AP^^BC^-\-PC^  +  AP^-^2TeGt.  BC-PC.   (Ax.  2) 

But  PB^  +  ^^  =0=  'ab\  and  Fc^  +  AP^  =  Ac\'  (347) 

.-.   AB^  =0  BC"  +  AC^  +  2  rect.  BC  •  PC.  Q.E.D. 

352.  Cor.  If  the  square  of  one  side  of  a  triangle  is  equiva- 
lent to  the  sum  or  the  difference  of  the  squares  of  the  other  two 
sides,  in  either  case  the  two  lesser  sides  are  at  right  angles  to 
each  other  (347,  350,  351). 

Exercise  471.  Show  that  the  difference  of  the  square  of  a  line  and 
the  square  of  its  projection  on  another  line,  is  equivalent  to  the  square 
of  the  difference  of  the  perpendiculars  that  intercept  the  projection. 

472.  In  the  diagram  for  Prop.  VI.,  show  that  AB^ -\- W]^  - Z€^ 

+  2  rect.  AB'EB,  and  UD^  +  AD^  =  Ajf  -  2  rect.  CD  .  {AE -  CD). 

« 

473.  Show,  by  means  of  the  foregoing  exercise,  that 

3^+  BV"-iUD''+ XZ)^)<>2(rect..4  B(AB-  AE)  +  rect.  CD{AE-  CD)). 


182  PLANE   GEOMETRY.  — BOOK   V. 

Proposition  XIV.     Theorem. 

353.  The  sum  of  the  squares  of  any  two  sides  of  a 
triangle  is  equivalent  to  twice  the  square  of  the 
median  to  the  third  side  plus  twice  the  square  of  half 
this  side. 


Given :  In  triangle  ABC,  AD  the  median  drawn  to  bO; 
To  Prove :  AB^  +  AC^  is  equivalent  to  2  AD^  +  2  'bd^. 

Draw  AP  ±  to  BC. 
1°.  li  AB  =  AC,  then  AP  coincides  with  AD,       (74) 
AB^ ^ Bif  4-  Aif,  and  Ajf  =o=  dc^  +  ad^ ;        (347) 
.-.  ab'-\- Ac"^  =0=2  BD^-{- 2  Ad'\  q.e.d.  (Ax.  2) 
(since  DC  =  BD.) 
2°.  If  AB>  AC,  Z  ADB  is  obtuse  and  Z  ADC,  acute  ; 

.:   AB'^AD^ -\- BD^ -\-2BD  •  DP,  (351) 

and  AC^^AD^  +  DC^  —  2 DC  -dp;  (350) 

.-.  AB^  + ag^^2ad~ +  2bd\     q.e.d.   (Ax.  2) 


Exercise  474.  In  the  diagram  for  Prop.  XII.,  left-hand  figure,  if  BP' 
be  drawn  perpendicular  to  ^O,  sliow  that  rect.  ^  C  •  P'C^^rect.BG-  PC. 

475.  In  the  same  diagram,  if  AB^  ^  370'^+  3  PO^  in  what  way  does 
P  divide  BG? 

476.  From  the  diagraip  for  Exercise  474  deduce  a  proof  of  the 
theorem :  The  altitudes  of  a  triangle  are  inversely  proportional  to  the 
sides  to  which  they  are  drawn. 


QUADRILA  TERALS.  183 

Proposition  XV.     Theorem. 

354.  The  rectangle  of  any  two  sides  of  a  triangle  is 
equivalent  to  the  rectangle  contained  by  the  altitude 
upon  the  third  side  and  the  diameter  of  the  cir- 
cumscribed circle. 


Given :  Triangle  ABC  inscribed  in  circle  ABBC,  AB  a  diamfeter, 
and  AP  perpendicular  to  BC ; 

To  Prove :  Eect.  AB  -  AC  i^  equivalent  to  rect.  AP  -  AB. 

Join  BC. 

Since  ACB  is  a  semicircle,  (Hyp.) 

rt.  Z  ACB  =  rt.  Z  APB;  (267) 

2i[so  Zb=Zb;  (266) 

.-.  A  APB  is  similar  to  A  ACB  ;  (288) 

.'.   AB'.AP  =  AB:AC ;  (284) 

.-.  rect.  AB  •  AC  ^  rect.  AP  •  AB.   q.e.d.   (321) 


Exercise  477.  If  ^O  is  a  diagonal  of  a  parallelogram  ABCD, 
having  /LA  equal  to  an  angle  of  an  equilateral  triangle,  show  that 
ZU^^ZB'^  +  W:!^  +  rect.  AB-BG. 

478.  If  AC  is  a  diagonal  of  a  parallelogram  ABCD,  having  ZA 
equal  to  twice  an  angle  of  an  equilateral  triangle,  show  that  AjO^ 
<^AB'^  +  BC'^  -  rect.  ABBC. 

479.  The  square  of  the  base  of  an  isosceles  triangle  is  equivalent  to 
twice  the  rectangle  contained  by  either  of  the  arms  and  the  projection 
of  the  base  upon  that  side. 


184  PLANE   GEOMETRY.  — BOOK    V. 

Proposition  XVI.     Theorem. 

355.  The  rectangle  of  any  two  sides  of  a  triangle  is 
equivalent  to  the  rectangle  of  the  segments  ijvade  on 
the  third  side  hy  the  bisector  of  the  opposite  angle, 
plus  the  square  of  the  bisector. 


Given:  In  triangle  ABC,  AD  tlie  bisector  of  angled,  cutting 
BG'm  D ; 

To  Prove ':  rect.  AB  -  ACi^  equivalent  to  rect.  BD  -  DC  +  AD^. 

Describe  O  ABEC  about  A  ABC;         '        (185) 
produce  AD  to  meet  circumference  in  £,  and  join  EC. 

Since  Z  BAD  —Z.  CAE,  (Hyp.) 

SindZB=:ZE,  (266) 

A  ABD  is  similar  to  A  AEC;  (287) 

.:  AB:  AD  =AE:AC;  (284) 

.-.  rect.  AB  .  ^C'=o=  rect.  AD  •  AE ;  (321) 

i.e.,  rect.  AB  -  AC=^  rect.  AD  •  {AD  +  DE)  ; 

i.e.,  rect.  AB  •  AC^  AD^  +  rect.  AD  •  DE ;  (335) 

i.e.,  rect.  AB  •  .1C=<>  rect.  BD  -  DC  -\-  AD^.  q.e.d.   (301) 

Exercise  480.     The  sum  of  the  squares  of  the  sides  of  any  paral- 
lelogram is  equivalent  to  the  sum  of  the  squares  of  the  diagonals. 

481.  A  median  divides  a  triangle  into  two  equivalent  triangles. 

482.  Three  times  the  sum  of  the  squares  of  the  sides  of  a  triangle 
is  equivalent  to  four  times  the  sum  of  the  squares  of  the  medians. 


QUADRILA  TERALS.  185 

Proposition  XVII.     Theorem. 

356.  If  two  chords  intersect,  the  rectangle  of  the 
segments  of  the  one  is  equivalent  to  the  rectangle  of 
the  segments  of  the  other. 

(See  diagram  for  Prop.  XXL,  Book  IV.) 

Since  OA  :  OD  =  OC:  OB,  (301) 

rect.  OA'OB  :o  rect.  OC  •  OD.     q.e.d.     (321) 


Proposition  XVIII.     Theorem. 

357.  If  from  the  same  point  a  tangent  and  a 
secant  he  drawn  to  a  circle,  the  square  of  the  tan- 
gent is  equivalent  to  the  rectangle  of  the  secant  and 
its  external  segment. 

(See  diagram  for  Prop.  XXIII.,  Book  IV.) 

Since  OA:OC=OC:  OB,  (303) 

OC^^rect.  OA  '  OB.  Q.E.D.     (322) 

Scholium.  The  two  foregoing  propositions  enunciate 
the  last  three  of  Book  IV.  from  a  different  point  of  view. 
In  these  and  similar  theorems,  we  may  substitute  product 
for  rectangle  when  regard  is  had  to  the  numerical  measures 
of  the  lines  concerned. 

Exercise  483.  In  any  triangle  ABG^  if  the  altitudes  BD^  CE, 
be  drawn  to  AC,  AB,  respectively,  show  that  BU^ <^ rect.  AB  -  BE  -^ 
rect.^C-  CD. 

484.  If  ABC  is  a  scalene  triangle, 

'AB^  +  AJ)'^-\-BV'^>AB-AC-\-AB-BC-^AC'BC. 

485.  The  square  of  the  median  to  the  hypotenuse  of  a  right  tri- 
angle is  equivalent  to  one  fourth  of  the  square  of  the  hypotenuse. 


186  PLANE   GEOMETRY.  — BOOK   V. 

EXERCISES. 

THEOREMS. 

486.  A  parallelogram  is  divided  by  its  diagonals  into  four  equiva- 
lent triangles. 

487.  If  two  triangles  have  two  sides  of  the  one  severally  equal  to 
two  sides  of  the  other,  and  the  included  angles  supplementary,  the 
triangles  are  equivalent. 

488.  If  any  point  within  a  parallelogram  be  joined  with  the  ver- 
tices, the  sums  of  the  opposite  pairs  of  triangles  are  equivalent. 

489.  If  through  any  point  in  a  diagonal  of  a  parallelogram  parallels 
to  the  sides  be  drawn,  of  the  four  parallelograms  thus  formed,  the 
two  through  which  the  diagonal  does  not  pass  are  equivalent. 

490.  A  line  joining  the  mid  points  of  its  bases  bisects  a  trapezoid. 

491.  If  the  mid  points  of  the  sides  of  a  quadrilateral  be  joined  in 
order,  a  parallelogram  is  formed  equivalent  to  one  half  the  quadri- 
lateral. 

492.  The  lines  joining  the  mid  point  of  a  diagonal  of  a  quadrilateral 
with  the  opposite  vertices,  cut  off  one  half  the  quadrilateral. 

493.  The  sum  of  the  squares  of  the  sides  of  any  quadrilateral  is 
equivalent  to  the  sum  of  the  squares  of  the  diagonals,  and  four  times 
the  square  of  the  line  that  joins  their  mid  points. 

494.  If  from  any  point  P  in  the  production  of  ^O,  a  diagonal  of  a 
parallelogram  ABCD^  lines  PB,  PD,  be  drawn,  the  triangles  PBC, 
PD  C,  will  be  equivalent. 

495.  If  from  a  point  P  without  a  parallelogram  ABCD,  lines  PA, 
PB,  PC,  PD,  be  drawn,  then  A  PAB  -  A  PCD  is  equivalent  to  one 
half  the  parallelogram. 

496.  Of  the  four  triangles  formed  by  drawing  the  diagonals  of  a 
trapezoid,  (1)  those  having  as  bases  the  nonparallel  sides  are  equiva- 
lent ;  (2)  those  having  as  bases  the  parallel  sides  are  as  the  squares  of 
those  sides. 

497.  If  two  triangles  have  a  common  angle  and  equal  areas,  the 
sides  containing  the  common  angle  are  inversely  proportional. 

498.  If  through  its  vertices  lines  be  drawn  parallel  to  the  diagonals 
of  any  quadrilateral,  the  figure  formed  will  be  a  parallelogram  of  twice 
the  area  of  the  quadrilateral. 


CONS  TR  UC  TIONS.  1 87 

499.  In  the  diameter  of  a  circle  two  points  are  taken  equally  dis- 
tant from  the  center ;  if  through  one  of  these  any  chord  be  drawn, 
and  its  extremities  joined  with  the  other  point,  the  sum  of  the  squares 
of  the  triangle  formed  is  constant. 

500.  The  sum  of  the  squares  of  the  four  segments  of  any  two 
chords  that  intersect  at  right  angles  is  constant. 

501.  The  rectangle  of  the  segments  of  chords  intersecting  at  a 
given  distance  from  the  center  is  constant. 

502.  The  square  inscribed  in  a  semicircle  is  to  that  inscribed  in 
the  circle  as  2  is  to  5. 

503.  If  one  side  of  a  triangle  is  lengthened,  and  another  shortened 
by  the  same  length,  the  line  joining  the  points  of  section  is  divided 
by  the  base  in  the  inverse  proportion  of  the  sides. 

504.  If  from  the  same  point  a  secant  and  two  tangents  be  drawn, 
the  secant  will  be  divided  harmonically  by  the  circumference  and  the 
chord  joining  the  points  of  contact. 


CONSTRUCTIONS. 

358.  To  construct  a  triangle  equivalent  to  a  given  polygon 
AB-'-F.  E 

Join  the  extremities  of  any  two  yyj  ^ 

adjacent  sides,  AF,  FE,  by  AE ;  and      "^  /  /  \ 

through  i^  draw  i^x  parallel  to  ^^,  to      /  y    /  ) 

meet  BA  produced  in  X.     Join  EX.        /  \  /  / 

We  have  now  a  polygon  BCDEX,  hav-       A^ / 

ing  one  side  fewer  than  the  given 

polygon,  and  equivalent  to  it,  since  A  XEA  ^  A  FEA  (332). 

By  now  joining  BE,  and  proceeding  as  before,  we  obtain 
an  equivalent  polygon  having  one  side  fewer  than  BCDEX, 
and  two  fewer  than  the  given  polygon.  Continuing  the 
process,  we  evidently  must  at  last  obtain  a  triangle  equiva- 
lent to  the  given  polygon. 


188  PLANE  GEOMETRY.  — BOOK   V, 

359.  To  construct  a  parallelogram  equivalent  to  a  given 

triangle  ABC,  and  having  an   angle  z y        a 

equal  to  a  given  ayigle  D.  V         \       7\ 

Through  A   draw  AZ  parallel   to  \         \/     \ 

BC,  and  through  X,  the  mid  point  of  \     \        A  \ 

BC,  dmw  XY,m2ikmg  Zbxy=Zd,  A     \  /   \  \ 

and  meeting  AZ  in  Y.     Through  B  ^ ji q 

draw  BZ  parallel  to  xr  to  meet  AZ 

in  z.     Then  xz,  which  is  equivalent  to  A^^C  (331),  is  the 

parallelogram  required. 

Scholium.     If  the  given  angle  z>  is  a  right  angle,  then 
XZ  will  be  a  rectangle. 

360.  To  construct  a  square  equivalent  to  a  given  rectangle  A  C. 
Produce  AB  to  E,  so  that  BE  =  BC, 

and  upon  AE  as  diameter  describe  a 
semicircumference  AXE.  Produce  BC 
to  meet  the  circumference  in  X.  Then 
-BX  is  a  side  of  the  required  square. 


For  since  AB  :BX  =  BX:  BE,  (298) 

BX^^AB  '  BE=o=AB  '  BC.  (322) 

Scholium.  By  means  of  the  three  foregoing  construc- 
tions, we  can  construct  a  square  equivalent  to  any  given 
polygon  by  first  transforming  the  given  polygon  into  a 
triangle,  then  the  triangle  into  a  rectangle,  and  finally  the 
rectangle  into  a  square. 

361.  To  construct  a  square  equivalent  to  the  sum  of  two 
given  squares.  q 

Draw  BC  ±  to  AB,  and  make  AB, 
BC,  respectively  equal  to  the  sides  of 
the  given  squares.     Join^C.     Then 

2  2  2 

AG  =c=AB   -{-  BC  . 


CONSTRUCTIONS.  189 

Scholium.  It  is  obvious  that,  by  a  continuation  of  the 
process,  we  can  obtain  the  side  of  a  square  equivalent  to 
the  sum  of  any  number  of  given  squares. 

362.  To  construct  a  square  equivalent  to  the  difference  of 
two  given  squares. 

Draw  B3I  ±  to  BN,  and  on  BN  lay  off  BA  =  the  side  of 
the  lesser  square.     From  A  as  cen- 
ter,  with  radius  AX  =  the  side  of  ^^^-.^ 

the  greater  square,  describe  an  arc  '\lr 

cutting  BM  in  X.      Then  BX  is  a  ...•••■■'  \ 

side  of  the  required  square.     For  ....•••""  ^ 

BX^^^^AX^'-AB^        (347)      ^  ^  i^ 


363.  To  construct  a  square  that  shall  be  any  given  part  of 
a  given  square. 

That  is,  L  being  the  side  of  the  given  square,  we  have  to 
find  a  line  X,  such  that 

X:L  =  ^m:Vn,  or  x'  =  -L^. 

n 

Employing  the  construction  of  Art.  362,  we  take  BA=  ^^~^x 

and  AX  =  — '^^L  ;  that  is,  we  divide  L  into  2n  equal  parts 
2n 

(207),  then  take  m  —  n  and  m-\-n  such  parts. 

\   2n  J  \   2n   J  n 

If,  for  example,  we  wish  to  find  the  side  of  a  square 
that  shall  be  equivalent  to  -y-  of  a  given  square  whose  side 
is  a  line  L,  we  divide  L  into  12  equal  parts  (207),  take 
BA  =  (11  —  6),  or  5,  of  these  parts,  and  AX  =  (11  +  6),  or 
17,  of  these  parts.     Then 

BX^  =  ({iyL^-(^\yL'=l^l.L^,  or  X:i=Vll:  V6. 
If,  again,  we  wish  to  find  a  line  whose  numerical  meas- 
ure shall  be  Vl3,  supposing  L  to  be  the  linear  unit,  we 


190  PLANE   GEOMETRY.  — BOOK    V. 

1S  —  1  1S4-1 

take  BA  =^  — -L,  or  6  L,  and  AX  =       "^    L,  or  7  L.    Then, 

2  A  

since  B^  =  7^L^  —  6^L^  =  13l^,  we  have  BX  =  Vl3z  ;  ^.e., 
the  numerical  measure  of  BX  is  Vl3. 

364.  To  construct  a  polygon  P  similar  to  a  given  polygon 
Q,  so  that  we  shall  have 

P  :  Q  =  m  :  n. 

Let  L  be  any  side  of  Q ;  then  by  Art.  363  find  a  line  X 
such  that 

X'.L  =  Vm :  Vw. 

Upon  X  construct  (310)  a  polygon  P  similar  to  the  given 
polygon  Q.  Then  P  is  the  polygon  required;  since  (344) 
P:  Qz={^my:  {^ny  =  m:n. 

365.  To  construct  a  polygon  similar  to  each  of  two  given 
polygons,  and  equivalent  to  their  sum  or  difference. 

We  obtain  an  homologous  side  of  the  required  polygon  by 
proceeding  with  two  homologous  sides  of  the  given  poly- 
gons as  we  did  with  the  sides  of  the  given  squares  in  Art. 
361,  when  we  wish  to  find  a  sura ;  and  as  jve  did  with  the 
sides  of  the  given  squares  in  Art.  362,  when  we  wish  to 
find  a  difference.  Upon  the  base  thus  obtained  we  then 
construct  (310)  a  polygon  similar  to  the  given  polygons. 

366.  To  construct  a  rectangle  equivalent  to  a  given  square, 
and  having  the  sum  of  two  adjacent  sides  equal  to  a  given 
line  AB. 

Upon  ^^  as  diameter,  describe  a  semicircumference  ADB. 
At  A  draw  AC  A.  to  AB,  and  equal  to 
the  side  of  the  given  square.  Through 
0  draw  CD  II  to  AB  to  meet  ADB  in  D. 
From  D  draw  DX  ±  to  AB  to  meet  AB 
in  X.  Then  AX  and  BX  are  respec- 
tively the  base  and  altitude  of  the  required  rectangle.  For 
rect.  AX'BX^  DX^  ^  ag\  (298) 


X  B 


NUMERICAL   APPLICATIONS. 


191 


367.    To  construct  a  rectangle  equivalent  to  a  given  square, 
and  having  the  difference  of  two  adjacent 
sides  equal  to  a  given  line  AB. 

Upon  AB  as  diameter,  describe  a  circum- 
ference AXBY.  At  A  draw  AC  A- to  AB, 
and  equal  to  the  side  of  the  given  square. 
From  c  draw  the  secant  CXY  through  the 
center  to  meet  the  circumference  in  x  and  Y.  Then  CX, 
CY  are,  respectively,  the  base  and  altitude  of  the  required 
rectangle.     For 


rect.  CY  -CX 


(357) 


368.  To  construct  a  polygoii  similar  to  a  given  polygon  P, 
and  equivaleyit  to  another  given  polygon  Q. 

Find  (360,  Scholium)  X  and  Y,  the  sides  of  squares  equiv- 
alent to  P  and  Q  respectively.  Then  (305)  find  a  fourth 
proportional  Z,  to  X,  F,  and  L,  the  base  of  P.  The  polygon 
constructed  on  z  as  base  and  similar  to  P  (310)  will  be  the 
required  polygon.     Proof  by  the  student. 


NUMERICAL  APPLICATIONS. 

1°.  To  compute  the  altitude  of  any  triangle  in  terms  of  its 

>:^ 
In  the  A  ABC,  let  a,  b,  c,  be  the  numerical 
measure  of  the  sides  opposite  A,  B,  c,  respec- 
tively, and  h  that  of  the  altitude  from  A. 

Of  the  A  B   and  c,  one  at  least,  say  C, 
must  be  acute ; 

A  h^=b^-Dc\' 
3d  c^  =  a^  +  52  -  2  o^  op  ; 
a^^b^-cK 


CD  = 


2a 


192  PLANE   GEOMETRY.  — BOOK  V. 

k' = b'  (^'  +  ^'  ~  "^'Y^  ^  ^'^'  ~  ^'''  ^  ^'  ~  ^'^' 

^(2ab-\-a^-\-b'-  c^)  (2  ab  -  g^  -  6^  +  c') 
4a2 

^](a-^by-cmc'-ia-by\ 
4.0? 

_{a-\-b  +  c){a  +  b  —  c){c-\-a  —  b){c—a-{-b) 
~  4.0? 

Let  a4-6  +  c  =  2s;  le.,  let  s  denote  half  the  perimeter; 
then  a-\-b  —  c  =  2(s  —  c)  ; 
c  +  a-6  =  2(s-6); 
c  —  a  +  6  =  2(s  — a). 
Hence,    simplifying    and    extracting    the    square    root, 
2 


h  =  -Vs(s  —  a)  (s  —  6)  (s  —  c). 
2°.   To  compute  the  medians  of  a  triangle  in  terms  of  the 


In  A  ABC,  denoting  the  values  of  the  sides  as  in  1°,  and 
the  median  from  Ahy  m,  ^ 

since  ft^  +  c"  =  2  m^  +  2  (i  a)  \        (353) 


3°.  To  compute  the  bisectors  of  a  triangle  in  terms  of  the 
sides. 

In  A  ABC,  denoting  the  values  of  the  sides  as  in  1°,  and 
the  bisector  from  A,  by  d  (see  diagram  for  Prop.  XVI.)  ; 

since  be  =  BD  X  DC -\-d^,  (355) 

d^=bc-BD  XDC.  (1°) 


NUMERICAL  APPLICATIONS.  193 


■r.     ,    BD         DC         Bn-{-DC  a  .r^rjns 

But  —  =    -  =  — — ^ =  — -- ;  (278) 

c         b  b.-\-  c         0  +  c 


6  +  c  b  +  c 

Substituting  these  values  in  1°,  and  simplifying, 

2        

^  =  7~, —  ^bcs(s  —  a), 
b  -he  ^  ^ 

4°.   To  compute  the  radius  of  the  circumscribed  circle  in  terms 
of  the  sides  of  a  triangle  ABC. 

In  A  ABC,  denoting  the  values  of  the  sides  as  in  1°,  and 
the  radius  by  R  (see  diagram  for  Prop.  XV.) ; 

since  bc  =  2Rx  AP,  (354) 

2     __^ 

and  AP  =  ^-^s{s-  a)  {s  -b){s-  c),  (1°) 


K 


a 

ahc 


4  Vs(s  —  a)  (s  —  b)  {s  —  c) 


5°.    To  compute  the  area  of  a  triangle  in  terms  of  its  sides. 

Denoting  the  values  of  the  sides  as  in  1°,  that  of  the  alti- 
tude by  h,  and  of  the  area  by  S, 

2 


since  8=1-  ah  (331),  and  7i  =  -  ^s{s-a)  (s-b)  (s-c),     (1°) 

a 


S=  Vs(s—  a)(s  —  b){s  —  c). 

Scholium.     If  the  triangle  is  equilateral,  i.e.,  if  a=b=c, 

a^ 
the  formula  reduces  to  *s  =  —  ^^3, 

6°.   To  compute  the  area  of  a  triangle  in  terms  of  the  sides 
and  the  radius  of  the  circumscribing  circle. 
Geom.  — 13 


194  PLANE   GEOMETRY.— BOOK    V. 

Denoting  the  values  of  the  sides,  altitude,  and  area  as 
in  5°, 

since  bG  =  2  R  -h,  (354) 

abc  =  2R'a-h  =  4:R'S; 

.    o  _abc 
~4^ 


s 


EXERCISES. 

QUESTIONS. 


505.  How  many  different  altitudes  can  each  of  the  following  figures 
have  :  An  equilateral  triangle  ?  An  isosceles  triangle  ?  A  scalene 
triangle  ?     A  square  ?     A  rectangle  ?    A  trapezoid  ?    A  trapezium  ? 

606.   A  side  of  an  equilateral  triangle  is  6.*    What  is  its»altitude  ? 

507.  An  arm  and  the  base  of  an  isosceles  triangle  are  18  and  16 
respectively.     What  are  its  altitudes  ? 

508.  The  area  of  a  triangle  is  180 ;  its  sides  are  30,  60,  and  40, 
respectively.     What  are  its  altitudes  ? 

509.  The  area  of  a  triangle  is  252  ;  its  altitudes  are  8,  12,  and  14, 
respectively.     What  are  its  sides  ? 

510.  The  sides  of  a  rectangle  are  65  and  32  respectively.  What 
are  its  area,  perimeter,  and  diagonal  ? 

511.  The  altitude  and  base  of  a  triangle  being  23  and  10  respec- 
tively, what  is  its  area  ? 

512.  The  area  of  a  triangle  is  221  sq.  ft.  ;  its  base  is  5|  yds.  What 
is  its  altitude  in  inches  ? 

513.  The  bases  of  two  parallelograms  are  15  and  16  respectively ; 
their  altitudes  are  8  and  10  respectively.  What  is  the  ratio  of  their 
areas? 

514.  Two  triangles  of  equal  areas  have  their  bases  26  in.  and  3  ft. 
respectively.     What  is  the  ratio  of  their  altitudes  ? 

*  Remember  that  the  given  abstract  numbers  are  numerical  measures. 


EXERCISES.  195 

515.  The  bases  of  a  trapezoid  are  23  in.  and  17  in.  respectively, 
its  altitude  being  2 \  ft.    What  is  its  area  ? 

616.   In  A  ABC,  AB  =  i2,  AG  =  34.    If  DE 

cut  off  AD  =  30  and  AE  =  15,  what  is  the  ratio 
of  A^^Oto  A^Z>^? 

517.  What  should  be  the  length  of  a  ladder 
such  that,  having  its  foot  15  ft.  from  the  wall, 
it  may  reach  a  window  20  ft.  from  the  ground  ? 

518.  Two  chords  intersect  so  that  the  segments  of  one  are  12 
and  7  respectively.  If  a  segment  of  the  other  is  10,  what  is  its 
second  segment  ? 

519.  What  are  the  altitudes  of  a  triangle  whose  sides  are  12,  15,  9, 
respectively  ? 

520.  What  are  the  medians  of  the  same  triangle  ? 

521.  What  are  the  bisectors  of  the  angles  of  the  same  triangle  ? 

522.  What  is  the  radius  of  the  circle  circumscribing  the  same 
triangle  ? 

528.  What  is  the  area  of  the  same  triangle  ? 

524.  What  is  the  area  of  a  triangle  whose  sides  are  6,  5,  5,  respec- 
tively ? 

525.  What  is  the  area  of  an  equilateral  triangle  whose  side  is  3  ? 

526.  What  is  the  area  of  an  equilateral  triangle  whose  altitude 
is  11  ? 

527.  The  sides  of  a  right  triangle  are  25,  24,  7.  What  are  its 
medians  and  its  altitude  upon  the  hypotenuse  ? 

528.  From  the  same  point  a  tangent  and  a  secant  being  drawn,  if 
the  secant  and  its  external  segment  are  as  27  to  3,  what  is  the  length 
of  the  tangent  ? 

529.  If  from  the  point  just  referred  to  a  second  secant  be  drawn,  so 
that  its  external  segment  is  8,  what  will  be  the  length  of  the  secant  ? 

530.  Two  secants  drawn  from  the  same  point  have  external  seg- 
ments of  5  and  3  respectively.  If  the  first  secant  is  27,  what  are  the 
internal  segments  ? 


196  PLANE   GEOMETRY.  — BOOK    V. 


PROBLEMS. 

531.  Construct  an  isosceles  triangle  on  the  same  base  as  a  given 
triangle,  and  equivalent  to  it.    • 

532.  Construct  a  right  isosceles  triangle  equivalent  to  a  given 
square. 

533.  Construct  a  parallelogram  having  a  given  angle  upon  the  same 
base  as  a  given  square,  and  equivalent  to  it. 

634.  Divide  a  given  line  into  two  segments  such  that  their  squares 
shall  be  as  7  is  to  5. 

535.  Bisect  a  given  parallelogram  (1)  by  a  line  passing  through  a 
given  point ;  (2)  by  a  perpendicular  to  a  side  ;  (3)  by  a  line  parallel 
to  a  given  line. 

536.  Bisect  a  given  triangle  by  a  line 
drawn  through  a  given  point  P  in  one  of  the 
sides. 

537.  Cut  off  one  nth  of  a  triangle  by  a 
line  drawn  through  a  given  point  in  one  of 
the  sides. 

538.  Bisect  a  quadrilateral  by  a  line  drawn 
through  one  of  the  vertices. 

539.  Cut  off  from  a  quadrilateral  one  nth 
part  by  a  line  drawn  through  one  of  the  vertices. 

540.  Bisect  a  triangle  by  a  line  parallel  to  the  base. 

541.  Bisect  a  triangle  by  a  line  perpendicular  to  the  base. 

542.  Find  a  point  within  a  triangle  such  that  lines  joining  the 
point  with  the  vertices  shall  divide  the  triangle  into  three  equivalent 
triangles. 

543.  Bisect  a  trapezoid  by  a  line  drawn  parallel  to  the  bases. 

544.  Bisect  a  trapezoid  by  a  line  drawn  through  a  given  point  in  one 
of  the  bases. 

545.  Construct  a  triangle  equivalent  to  a  given  triangle,  and  having 
one  side  equal  to  a  given  line. 

546.  Construct  a  right  triangle  equivalent  to  a  given  triangle,  and 
having  one  arm  of  a  given  length. 


EXERCISES.  197 

547.  Construct  a  right  triangle  equivalent  to  a  given  triangle,  and 
having  its  hypotenuse  of  a  given  length. 

548.  Construct  an  isosceles  triangle  equivalent  to  a  given  triangle, 
and  having  its  arms  of  a  given  length. 

549.  Construct  an  isosceles  triangle  equivalent  to  a  given  triangle, 
and  having  its  base  of  a  given  length. 

550.  Construct  an  equilateral  triangle  equivalent  to  a  given  triangle. 
651.  Construct  an  equilateral  triangle  equivalent  to  a  given  square. 

552.  Construct  a  triangle  similar  to  each  of  two  given  similar  tri- 
angles, and  equivalent  to  their  sum. 

553.  Construct  a  triangle  similar  to  each  of  two  given  similar  tri- 
angles, and  equivalent  to  their  difference. 

554.  Construct  a  square  that  shall  be  to  a  given  triangle  as  5  is  to  3. 

555.  Construct  an  equilateral  triangle  that  shall  be  to  a  given  square 
as  7  is  to  5. 


Book   VL 
regular  plane  figures. 

regular  polygons. 

369.  A  polygon  of  five  sides  is  called  a  pentagon ;  one  of 
six  sides  is  called  a  hexagon;  of  seven  sides,  a  heptagon; 
of  eight  sides,  an  octagon;  of  fifteen  sides,  a  pentadecagon ; 
and  so  on. 

370.  A  regular  polygon  is  both  equilateral  and  equiangular. 

PiioposiTiox  I.     Theorem. 

371.  A  regular  polygon  may  he  divided  into  as 
many  equal  isosceles  triangles  as  the  polygon  has 
sides. 


E  D 


Given :  A  regular  polygon  AB  -"  F,  or  P,  having  n  sides ; 
To  Prove :  P  may  be  divided  into  w  equal  isosceles  triangles. 

Bisect  A  A  and  B  by  ^0,  BO.  (81) 

A  A  and  B  are  each  <  a  st.  Z  ; 
.-.  ^Z  A  -\-}Zb  <  Si  st  Z; 
.*.  AO,  BO,  must  meet  in  some  point  0.        (114) 
198 


REGULAR   POLYGONS.  199 

Join  0  Avitli  C,  D,  E,  F ; 
then  since  Z  OAB  =  Z  OB  A,  (Ax.  7) 

A  OAB  is  isosceles.  (65) 

Since  AB  =  BC  (Hyp.),  OB  is  common, 

and  Z  OB  A  =  ZOBC,  (Const.) 

AOBa  =  AOAB.  (66) 

In  the  same  way  it  may  be  shown  that 
AOCD  =  isos.  A  OBC,  A  OAF  =  isos.  OAB,  etc.       q.e.d. 

372.  CoR.  1.  The  bisectors  of  any  two  angles  of  a  regular 
polygon  determine  by  their  intersection  a  point  equidistant  from 
the  vertices  of  the  polygon. 

For  the  lines  drawn  from  the  vertices  to  that  point  are 
the  sides  of  equal  isosceles  triangles  (371). 

373.  Cor.  2.  The  point  that  is  equidistant  from  the  vertices, 
is  also  equidistant  from  the  sides,  of  the  pjolygoyi  (101). 

For  it  is  in  the  bisectors  of  all  the  angles  of  the  polygon. 

374.  Cor.  3.  A  circle  may  be  circumscribed  about,  or  in- 
scribed in,  any  regular  polygon,  and  both  circles  have  the 
same  center. 

For  taking  the  intersection  of  the  bisectors  of  any  two 
angles  as  center,  the  circumference  described  through  one 
vertex  will  pass  through  all  (372),  and  that  described  tan- 
gent to  one  side  will  be  tangent  to  all  (373). 

375.  Definitiox.  The  center  of  a  regular  polygon  is  the 
common  center  of  the  inscribed  and  circumscribed  circles. 

376.  Definition.  The  radius  of  a  regular  polygon  is 
that  of  the  circumscribed  circle. 

377.  Definition.  The  opothem  of  a  regular  polygon  is 
the  radius  of  the  inscribed  circle;  i.e.,  the  distance  from 
the  center  to  any  side. 


200  •      PLANE   GEOMETRY.  — BOOK    VL 

378.  Definition.  An  angle  at  the  center  of  a  regular 
polygon  is  the  angle  formed  by  the  radii  drawn  to  the 
extremities  of  any  side. 

379.  CoR.  4.  //  the  number  of  sides  is  n,  each  angle  at 
the  center  =  2  st.  A-i-n. 


Proposition  II.     Theorem. 

380.  An  equilateral  polygon  Uiscribed  in  a  circle 
is  regular. 

E^ — -^  D 


Given:  An  equilateral  polygon  P  inscribed  in  (DACE; 
To  Prove :  P  is  a  regular  polygon. 

Since  chd.  AB  =  chd.  PC  =  chd.  CD,  etc.,         (Hyp.) 

arc  AB  =  arc  PC  =  arc  CD,  etc. ;  (174) 

.-.  arc  ABC  =  arc  BCD  =  arc  CDE,  etc. ;      (Ax.  2) 

.-.  ZBz=Zc  =  Ad,  etc.,  (266) 

(being  inscribed  in  equal  segments ;) 

.*.  P,  being  equilateral  and  equiangular,  is  regular. 

Q.E.D.      (370) 

381.  Cor.  1.  If  a  circumference  he  divided  into  n  equal 
arcs  (n  being  >  2),  the  chords  subtending  these  arcs  ivillform 
a  regular  polygon  of  n  sides. 

382.  Cor.  2.  If  the  arcs  subtended  by  the  sides  of  a  regu- 
lar polygon  o/n  sides  be  bisected,  the  chords  drawn  to  subtend 
these  arcs  willforyn  a  regular  polygon  of  2n  sides. 


REGULAR   POLYGONS.  201 

Proposition  III.     Theorem. 

383.  Regular  polygons  of  the  same  number  of  sides 
are  similar. 


Given:  Two  regular  polygons,  ABC  or  P,  and  A'b'c'  or  P', 
each  of  n  sides ; 

To  Prove:  P  is  similar  to  P'. 

Since  P  and  p'  have  each  n  angles,  (Hyp.) 

each  Z  of  P  and  p'  =  '-^^^  st.  A  resp. ;  (127) 

n 

.'.  P  and  7^'  are  mutually  equiangular. 
Since  P  and  P'  are  each  equilateral,         (HyP-) 
AB  :BC=  A^B^:B'& ; 
similarly,  all  the  sides  about  the  equal  A  are  proportional ; 
.*.  P  is  similar  to  P'.  q.e.d.     (284) 

384  Cor.  The  perimeters  of  regular  polygons  of  the  same 
number  of  sides  are  as  their  radii,  or  apothems;  and  their 
areas  are  as  the  squares  of  these  lines. 

For  the  radii  and  apothems  are  homologous  lines ;  hence 
the  perimeters  are  proportional  to  them  (296),  and  the 
areas  are  proportional  to  the  squares  of  those  lines 
(344). 

ExERGiSE  556.  The  ratio  of  an  interior  angle  of  a  regular  polygon 
of  n  sides  to  an  interior  angle  of  a  regular  polygon  of  double  the  num- 
ber of  sides,  is  expressed  by  n  —  2  :  ?i  —  1. 


202 


PLANE    GEOMETRY.  — BOOK    VI. 


Propositiox  IV.     Theorem. 

385.  A  regular  polygon  being  inscribed  in  a  circle, 

a  similar  polygon  may  be  circumscribed  about  the 

circle, 

L K 


M 


^.6^ 


N 


Given :  A  regular  polygon  AB  •-•  F,  or  P,  inscribed  in  O  ACE ; 
To  Prove:  A  similar  polygon  can  be  circumscribed  about  ACE. 

Through  each  vertex  A,  B,  C,  etc.,  draw  tangents  NG,  GH, 
HK,  etc. 

Since  AB  =  BC  =  CD,  etc.,  (Hyp.) 

Zgab  =  Z.  gba  =  Aubc  =  A  hcb,  etc.,       (269) 
(being  formed  by  tangents  and  chords  of  equal  arcs.) 
Since  each  chord  is  less  than  a  diameter, 
each  of  these  equal  A  is  less  than  a  rt.  Z ;      (267) 
.*.  the  tangents  through  adjacent  vertices  will  meet.     (114) 
Let  them  meet  in  G,  //,  K,  L,  etc. 
Since  isos.  Agab  =  isos.  A IIBC  =  isos.  A  KCD,  etc.,   (63) 
Zg  =  Zr  =  Zk=  etc. ;  (70) 

then  GHK  '••  JV  is  equiangular. 
Also,  GA  =  GB  =  HB  =z  lie  =  etc.,  (70) 

GH  =  HK  =  KL  =  etc.  ;  (Ax.  6) 

.-.  GHK"'N  is  equilateral,  with  the  same  no.  of  sides  as  P ; 
.'.  GHK  •••  iV^  is  a  regular  polygon  similar  to  P.  q.e.d.   (370) 

386.  CoR.  If  a  circumference  he  divided  into  n  equal  arcs 
(n  being  >  2),  the  tangents  drawn  at  the  points  of  division  ivill 
form  a  regular  circumscribed  polygon  of  n  sides. 


REGULAR  POLYGONS.  203 

Proposition  V.     Theorem. 

387.  Any  regular  polygon  is  equivalent  to  the  rec- 
tangle of  its  apothem  and  half  its  perimeter. 


A      R     B 

Given :  OB,  the  apothem  of  a  regular  polygon  AB  -•'  f,ot  p,  of 
n  sides ; 

To  Prove :  P  is  equivalent  to  rectangle  ^n  AB  -  OR. 

Draw  AO,  BO. 

Since  A  OAB  ^i  rect.  AB  •  OR,  (331) 

and  P  ^  ?i  A  OAB,  (371) 

P  =0=  i  n  rect.  AB  -  OR^  rect.  ^n  AB  -  OR.       q.e.d. 

Scholium.   The  theorem  may  also  be  stated  thus : 

The  area  of  a  regular  polygon  is  measured  by  one  half  the 
product  of  its  apothem  and  perimeter. 

Exercise  557.  In  the  diagram  for  Prop.  IV.,  if  GR  be  the  perpen- 
dicular from  Q-  to  AB,  the  number  of  sides  being  n,  show  that  GR^  is 
4  n  times  the  difference  of  the  sums  of  the  squares  of  the  sides  of  the 
circumscribed  and  inscribed  polygons. 

558.  In  the  same  diagram,  n  being  the  number  of  sides,  each  angle 
formed  like  GAB  by  an  interior  and  an  exterior  side,  is  -  of  a  straight 
angle.  ^ 

659.  In  the  diagram  for 'Prop.  V.,  if  a  parallel  to  AB  be  drawn 
through  the  mid  point  of  2?C,  what  lines  will  it  bisect  ? 

560.  In  the  same  diagram,  if  the  mid  points  of  the  radii  be  joined, 
what  figure  will  be  formed,  and  what  ratio  will  its  area  have  to  that 
otPf 


204  PLANE   GEOMETRY.  — BOOK   VI. 

Proposition  VI.     Theorem. 

388.  As  the  miwiber  of  sides  of  a  regular  in- 
scribed polygon  indefinitely  increases,  the  apothem 
increases  towards  the  radius  as  its  limit. 


Given :  AB,  a  side,  and  OP,  the  apothem,  of  a  regular  polygon 
of  n  sides  inscribed  in  a  circle  whose  radius  is  OA  ; 

To  Prove :  As  n  increases,  OP  increases  towards  OA  as  limit. 

Since  OP  is  ±  to  AB,  (377) 

OpVoI'-Zp'.  (347) 

Now  as  n  increases,  AB,  and  therefore  AP,  decreases  (181), 
and  when  n  becomes  indefinitely  great,  AP  becomes  in- 
definitely small,  while  OA,  the  radius,  is  constant ; 

.♦.  'op''  has  for  limit  Oa\'  (235) 

.-.  OP  has  for  limit  OA.  q.e.d. 

Exercise  561.  In  the  diagram  for  Prop.  VI.,  if  AO  be  produced, 
show  that  it  will  pass  through  a  vertex  of  the  polygon. 

562.  In  any  polygon  of  an  even  number  of  sides,  the  lines  joining 
opposite  vertices  are  diameters  of  the  circumscribed  circle. 

663,  In  the  diagram  for  Prop.  VI.,  if  the  inscribed  polygon  is  a 
regular  hexagon,  what  is  the  ratio  of  OP  to  AP  ? 

564.  In  the  same  diagram,  if  OB  be  joined,  and  PC,  PD,  be  drawn 
to  the  mid  points  of  OA,  OB,  resp.,  OCPD  will  be  a  rhombus,  unless 
Z  AOB  is  a  right  angle.    In  what  case  will  Z  AOB  be  a  right  angle  ? 

565.  The  area  of  the  regular  inscribed  hexagon  is  half  the  area  of 
the  circumscribed  equilateral  triangle. 


REGULAR   POLYGONS.  205 

Proposition  VII.     Theorem. 

389.  A  straight  line  is  the  least  of  all  the  lines  that 
terminate  in  two  given  points. 

^,^-<<- J2/  ^  p^^ ^>^ 


Given:  A  straight  line  AB,  and  another  line  AHG  -•'  Bf  termi- 
nating in  A,  B  ; 

To  Prove :  AB  is  less  than  AHG  •  •  •  j5. 

Join  alternate  points  AG,  GE,  EC. 
Since  AG  <  AH-\-HG,  GE  <  GF -{- FE,  EC<ED  +  DC,  (88) 
AGECB<AHGFEDCB.  (Ax.  4) 

By  continuing  the  process,  as  the  number  of  parts  is  al- 
ways decreasing,  we  shall  evidently  obtain  at  last  a  broken 
line  of  two  parts,  which  is  less  than  any  of  the.  preceding 
broken  lines,  and  greater  than  AB. 

Hence  AB  is  less  than  any  broken  line  terminating  in  A,  B. 

As  this  reasoning  holds  true,  no  matter  how  numerous  or 
how  small  the  parts  of  the  broken  line  may  be,  it  holds  true 
in  regard  to  curves  also,  which  are  the  limits  towards  which 
tend  the  broken  lines  formed  by  joining  points  of  the  curve 
taken  indefinitely  near  to  each  other. 

Hence  AB  is  less  than  any  other  line  terminating  in  A 
and  B.  Q.E.D. 

390.  Cor.  1.  An  arc  ACB  of  a  circle  is  less  than  any 
enveloping  line  ADB  that  terminates  in  the  same  points,  A 
and  B. 

Join  AB.    Of  all  the  lines  enveloping  the  ^-^     ~^\  ^ 

area  ^C-B,  there  must  be  a  least  one.    Now  -^V^- 

ADB  cannot  be  the  least,  for  drawing  EF      [l. 

tangent  to  A  CB  at  C,  we  have  AEFB  <  ADB,  ^^ 


206  PLANE    GEOMETRY.  — BOOK    VL 

since  EF  <  EDF.  In  the  same  way  it  can  be  shown  that  no 
other  line  than  ACB  can  be  the  least  line  enveloping  the 
area  ACB. 

391.  CoR.  2.  A  circumference  is  greater  than  the  peri'meter 
of  any  inscribed  polygon,  and  less  than  that  of  any  circum- 
scribed polygon. 


Proposition  VIII.     Theorem. 

392.  As  the  number  of  their  sides  indefinitely 
increases,  the  perimeters  of  regular  inscribed  and 
circumscribed  polygons  tend  towards  the  circumfer- 
ence as  their  common  limit,  and  their  areas  towards 
the  circle  as  their  common  limit. 

Given:  The  perimeters p,  p\  of  two  regular  polygons  P,  P',  of 
n  sides,  circumscribed  about,  and  inscribed  in,  a  circle  whose  circum- 
ference is  C  and  whose  area  is  s ; 

m    T^  {  ^°'  P  and  p'  each  tend  towards  C  as  limit. 

To  Jrrove :    ■{ 

{  2°.   P  and  P'  each  tend  towards  S  as  limit. 

Let  R  be  the  apothem  of  P  and  radius  of  P',  and  r  the 
apothem  of  P'. 

1°.  Since  p  12)' =B:r,  (384) 

P^zt  =  ^::zl.  (247) 

p  R 

Now  when  the  number  of  sides  becomes  indefinitely 
great,  r  approaches  its  limit  R  (388),  so  that  R  —  r  becomes 
indefinitely  small.    Hence  ^  ~  ^'  and  its  equal  ^~^  become 

indefinitely  small;  that  is,  p  —  p',  the  difference  of  the  perim- 
eters of  the  polygons,  becomes  less  than  any  assignable 
quantity,  p  being  not  much  greater   than  C.     Now  as    C 


REGULAR  POLYGONS.  207 

always  lies  between  p  and  ^'   (391),  its   difference   from 
either  must  be  less  thanp  —  p';  hence 

p  andjp'  have  each  for  limit  G.    q.e.d.    (235) 

2°.  Since  P:P'  =  R':r%  (384) 

Now   when   R^  —  ir^  becomes    indefinitely   small,   P  —  P' 
becomes  indefinitely  small;  and  since  always 

P  >  s,  hut  P'<s,  (Ax.  8) 

s  lies  always  between  p  and  p', 
(its  difference  from  either  being  less  than  P  —  P' ;) 

.-.  P  and  P'  have  each  for  limit  S.  q.e.d. 


Exercise  566.    The  apothem  of  an  inscribed  equilateral  triangle  is 
equal  to  half  the  radius  of  the  circle. 

567.  The  apothem  of  an  inscribed  regular  hexagon  is  equal  to  half 
the  side  of  the  inscribed  equilateral  triangle. 

568.  Any  arc  is  greater  than  its  chord,  but  less  than  the  sum  of  the 
tangents  drawn  from  a  point  to  its  extremities. 

569.  Every  equilateral  polygon  inscribed  in  a  circle  is  also  equi- 
angular. 

570.  Every  equiangular  polygon  inscribed  in  a  circle  is  also  equi- 
lateral, if  the  number  of  sides  is  odd. 

571.  Every  equilateral  polygon  circumscribed  about  a  circle  is  also 
equiangular,  if  the  number  of  sides  is  odd. 

572.  Every  equiangular  polygon   circumscribed   about  a  circle  is 
also  equilateral. 

573.  If  the  alternate  vertices  of  a  regular  hexagon  are  joined  by 
straight  lines,  show  that  the  figure  formed  is  a  regular  hexagon. 

674.  What  ratio  has  the  latter  figure  to  the  former  ? 


208  PLANE   GEOMETRY.  — BOOK  VI. 

Proposition   IX.     Theorem. 

393.  Circumfej^eiices    are    to  each    other    as   their 
radii. 


Given:  Two  circumferences,  C  C',  with  radii  R,  R' , 
To  Prove:  C  :  &  =  R  :  R'. 

Conceive  regular  polygons  of  n  sides  to  be  inscribed  in 
each  of  the  circumferences  (381),  and  by  continual  dou- 
bling (382),  the  number  of  sides  to  become  indefinitely 
great. 

Let  p,  p\  denote  the  variable  perimeters. 

Sincep:p'  =  i?:  J?',  (384) 

l=t-  (244) 

R       R'  ^ 

Now  the  limits  of  these  equal  variables  are  equal,  C 
being  the  limit  of  p  and  C'  of  p'  (392).     Hence 

--  =  ^.  Q.E.D.       (236) 

R      R^ 

394.  Cor.  1.  Circumferences  are  to  each  other  as  their 
diameters. 

YovC:C'  =  R'.R';  (393) 

.'.  C:&  =:2R:2r'  =  D:D'.  (253) 

395.  Cor.  2.  The  ratio  of  the  circumference  to  the  diame- 
ter is  constant. 

For  C':C' =  /):/)';  (394) 

.♦.  C:  J)  =  C' :  D\ 


REGULAR   POLYGONS.  209 

396.  Scholium.  The  numerical  value  of  this  constant 
ratio,  —  that  is,  the  number  showing  how  many  times  a  cir- 
cumference contains  its  diameter,  is  denoted  by  the  Greek 
letter  tt.  It  is  an  incommensurable  number,  but  its  value, 
as  will  presently  be  shown,  can  be  obtained  to  any  required 

degree  of  precision.    Since  —  =  - —  =  tt,  we  have  the  impor- 
tant relations, 

C  =  7r-Z>=  27ri?;  />=-,  R  =  ~- 
IT  2-^ 


Proposition   X.     Theorem. 

397.  A  circle  is  equivalent  to  one  half  the  rectan- 
gle contained  hy  its  radius  and  circumference. 


Given :  C,  the  circumference,  and  R,  the  radius,  of  a  circle  ABE  ; 
To  Prove :  Circle  ABE  \^  equivalent  to  \  rectangle  R  •  C. 

Let  r  denote  the  apothem,  and  p  the  perimeter,  of  a  regu- 
lar polygon  P  inscribed  in  ABE. 

Then  P  =c=  i  rect.  r  •  p.  (387) 

Conceive  the  number  of  sides  by  continual  duplication  to 

become   indefinitely  great.      Then  P   has   for   limit 

O  ABEf  and   rect.  r  •  j)  has   for  limit   rect.    R  -  C,  C 

being  the  limit  of  p  and  R  that  of  r;  (392) 

.-.  O^j^J^^i  rect.  R  '  c,  Q.E.D.     (236) 

(they  being  the  limits  of  variables  always  equal.) 

Geoin.  — 14 


210  PLANE   GEOMETRY.  — BOOK   VL 

« 

Scholium,  This  theorem  may  be  stated  under  the  form  : 
The  area  of  a  circle  is  measured  by  one  half  the  product  of 
its  radius  and  perimeter. 

398.  Cor.  1.  The  area  of  a  circle  is  equal  to  ir  times  the 
square  of  its  radius. 

For  denoting  the  area  of  the  circle  by  s, 

since  S^^R-  c,  (397),  and  C  =  27r  •  i?  (396), 

Sz=^R  X  27r-  R  =  7rR'\ 

399.  CoR.  2.  The  areas  of  circles  are  to  each  other  as  the 
squares  of  their  radii. 

For  S  =  TT  •  R",  and  S'^ir-  R';  (398) 

.-.   S:S'  =  7rR':7rR"=R':R". 

400.  Definition.  A  segment  of  a  circle  is  the  figure 
bounded  by  an  arc  and  its  chord. 

401.  Definition.  A  sector  of  a  circle  is  the  figure 
bounded  by  two  radii  and  their  intercepted  arc. 

402.  Definition.  Similar  segments  and  sectors  in  differ- 
ent circles  are  such  as  have  arcs  measuring  equal  angles 
at  the  center. 

403.  Cor.  3.  The  area  of  a  sector  is  measured  by  one 
half  the  product  of  its  radius  and  arc. 

For  the  sector  is  to  the  circle  as  the  arc  of  the  sector  is 
to  the  circumference. 

404.  Cor.  4.  Similar  sectors  are  as  the  squares  of  their 
radii. 

For  they  are  like  parts  of  their  respective  circles. 

Exercise  575.   What  figure  is  both  a  segment  and  a  sector  ? 

576.  The  area  of  one  circle  is  twice  that  of  another.  What  is  the 
ratio  of  their  radii  ? 

577.  The  radii  of  two  similar  segments  is  as  3  to  5.  What  is  the 
ratio  of  their  areas  ? 


REGULAR  POLYGONS. 


211 


Proposition  XI.     Theorem. 

405.  Similar  segments   are    to  each  other  as  the 

squares  of  their  radii. 

o  o' 


Given :  OA  =  R,  O'A^  =  R',  radii  of  similar 
To  Prove :  B  :  b'  =  R- :  R'^ 


8  B,  b' ; 


Since  Zo  =  Zo' 


(Hyp.) 
(402) 


sector  OABC  is  similar  to  sector  O'A'b'c'. 
Since  OA  =  OC,  O'A'  =  O'c',  and  Zo  =  Zo', 

A  0 AC  is  similar  to  A  O'A'c'. 

Since  sect.  OABC -.sect.  o'A'b'c'  =  r"^:  R'^, 

and  AOACiA  OWC'  =  OA^ :  O'A'^  =  R^ :  R'^, 

sect.  OABC :  A  OAC  =  sect.  oWb'c'  :  A  O'A'c' ;    (232'") 

.-.   OABC— OAC:  OAC  =  O'A'b'c' —  0W&:  O'A'c';  (247) 

.-.  seg.  B  :  seg.  B'  =:  A  OAC :  A  O'a'c'  =  R^ :  R'^.    q.e.d.    (244) 


(290) 
(404) 
(342) 


Exercise  578.  In  the  diagram  for  Prop.  XI.,  what  must  be  the 
ratio  of  OA  to  O'A'  if  segment  J5  is  f  of  segment  B'9 

579.  In  the  same  diagram,  if  the  segment  is  f  of  its  circle,  how 
many  degrees  are  there  in  arc  AC? 

580.  In  this  same  diagram,  if  A  were  joined  with  the  mid  point  D 
of  arc  AC,  and  Z  CAD  were  found  to  be  38°,  how  many  degrees  in 
ZAOC? 

581.  In  the  same  diagram,  if  a  line  ^F  were  drawn  perpendicular 
to  OA,  and  Z  CAF  were  found  to  be  41°,  how  many  degrees  in  arc 
AC?  How  many  in  the  angle  formed  by  lines  drawn  from  A  and  C 
to  any  point  in  arc  AC? 


212  PLANE   GEOMETRY.  — BOOK   VL 

DIVISION  OF  CIRCUMFERENCE. 

Proposition  XII.     Theorem. 

406.  A  circumference  can  be  divided  into  2,  4,  8, 
16,    "  equal  arcs. 


Given :  A  circumference  ACBD ; 

To  Prove :  ACBD  can  be  divided  into  2,  4,  8,  16,  •••equal  arcs. 

Find  the  center  of  ACBD.  (173) 

1°.  Through  0  draw  a  diameter  AB. 

AB  bisects  the  circumf.  into  the  equal  arcs  ACB,  ADB.   (169) 

2°.  Through  0  draw  a  diam.  CD  A.  to  AB.  (95) 

Since  the  angles  at  0  are  equal,  being  rt.  A,   (Const.) 

arc  ^C  =  arc  CB  —  arc  BD  —  arc  DA  =  \ACDB.    (257) 

3°.  Bisect  each  of  the  zi  at  0  by  diameters  EF,  GH.    (81) 

The  arcs ^^,J5:C',CG^,  etc.,  each  =  1  a  quad.  —\ACBD.  (Const.) 

By  successive  bisections  we  can  obtain,  in  the  same  way, 
the  16th,  32d,  •••2"th  part  of  ACBD,  and  hence  can  con- 
struct regular  polygons  of  4,  8,  16,  •••  2**  sides.        q.e.d. 

Exercise  582.  In  the  diagram  for  Prop.  XII.,  if  A  and  E  be 
joined  with  any  point  F  in  arc  AE,  how  many  degrees  in  the  angle 
thus  formed  ? 

583.  How  many  degrees  in  the  a,ngle  formed  by  joining  A  and  G 
with  any  point  in  arc  ADBG,  and  what  is  its  ratio  to  the  angle  found 
in  the  preceding  exercise  ? 


DIVISION  OF  CIRCUMFERENCE.  213 

Proposition  XIII.     Theorem. 

407.  J.   circumference   can  he  divided  into  S,   6, 
12,  "•  equal  arcs. 


Given:  A  circumference  ACBD ; 

To  Prove:  ACBD  can  be  divided  into  3,  6,  12,  •••  equal  arcs, 

Find  the  center  0  (173),  and  draw  any  diameter  AOB. 
1°.  From  B  draw  a  chord  BC  =  BO  (178),  and  join  CO. 

Since  CO  =  BO  =  BC,  (Const.) 

Zb  =  Zboc.  (68) 

But  Zb  is  inscribed,  and  Z  ^OC  is  at  the  center ; 

.-.  arc  ^C  =  2 arc ^C;  (268) 

i.e.,  SiTcAC  =  f  arc  BCA,  a  semicircumf. ; 
i.e.,  arc ^C  =  ^ACBD,  the  circumf. 


Since  arc  ^c  =  ^  arc  ^C, 
arc  5C  =  I  ACBD 


'}  <'°^ 


3°.    By  bisecting  arc  BC  again,  we  obtain  -^ACBD,  thence 

2V  ACBD,  and  so  on  to  any  arc  denoted  by  - — —ACBD. 

o  X.  Z 

Q.E.D. 

408.  Cor.  The  side  of  a  regular  inscribed  hexagon  is  equal 
to  the  radius  of  the  circle. 

For  BC,  which  subtends  ^ACBD,  is  equal  to  BO  (Const.). 
We  are  thus  enabled  to  construct  any  regular  polygon  of 
3x2"  sides. 


214  PLANE    GEOMETRY.  — BOOK    VI. 

Proposition   XIV.     Theorem. 

409.  A  circumference  can  he  divided  into  5,  10 y 
20,  ■■•  equal  arcs. 


Given :  A  circumference  A  CED  ; 

To  Prove:  AC  ED  can  be  divided  into  5,  10,  20,  •••  equal  arcs. 

Find  the  center  0  (173),  and  draw  any  radius  OA. 

1°.  Divide  OA  in  S,  so  that  OA  :  OB  =  OB  :  BA.      (308) 

From  A  draw  a  chord  AC  =  OB  (178),  and  draw  CBD,  to 

meet  the  circumference  in  D.     Draw  the  diameter  COE. 

Since   OA:  OB  =  OB:  BA,  > 

\  (Const.) 

and  AC=  OB,  and  OC  =  OA,  ) 

we  have  (1)  OA:  AC  =  AC:  BA, 

and  (2)  OC  :  AC  =  OB  :  BA. 

From  (1),  since  Z  ^  is  com.  to  both  A, 

A  BAC  is  similar  to  A  CA 0 ;  (290) 

.-.  Zacbot  ACD  =  Zaoc.  (289) 

From  (2),  since  the  sides  of  A  CAO  are  as  the  segments 

of  the  base  OA,  Z  ACB  =  Z  OCB ;  (279) 

.-.  arc  ED  =  arc  AD.  (257) 

But  arc  AD —  2  arc  AC,  (268) 

(since  Z  ACD  is  inscribed  and  Z  AOC  is  at  the  center;) 

.-.  arc  AD  4-  arc  DE  -f-  arc  AC  =^^  arc  AC ; 

.'.  arc  AD  —  \  arc  CDE  =  ^  ACED. 

2°.  Since  arc  AC=^\  arc  AD,  (1°) 

arc  AC  =  A^  arc  ACED, 


DIVISION  OF  CIRCUMFERENCE.  215 

3°.  By  bisecting  AC  we  obtain  -^^  ACED,  and  so  on.  Thus 
we  can  divide  the  circumference  ACED  into  5,  10, 
20,  •••  5  X  2"  equal  arcs.  q.e.d. 

410.  CoR.  Bij  taking  the  difference  of  two  arcs  respectively 
equal  to  ^  and  j\  of  the  circumfereyice,  since  i  —  yL  =  J^,  we 
can  find  an  arc  that  is  J^  of  the  circumference ;  and  thence, 
by  repeated  bisections,  arcs  that  are  -f^,  -^^,  -"  of  the  circum- 
ference. 

Till  the  beginning  of  this  century,  the  division  of  the  circumfer- 
ence by  means  of  the  straight  line  and  circle  could  be  effected  only  in 
the  cases  covered  by  the  preceding  propositions  ;  that  is,  the  circumfer- 
ence could  be  thus  divided  into  1  x  2»,  3x2",  5  x  2",  15  x  2",  equal 
arcs.  The  celebrated  geometer  Gauss  proved,  however,  that  the  cir- 
cumference can  be  divided  into  any  number  of  equal  arcs  expressed 
by  the  formula  2"  -}- 1  v^^hen  this  is  a  prime  number.  Thus,  as  we 
have  seen,  we  can  effect  the  division  into  2°  -f-  1  =  2,  2^  -|-  1  =  3, 
22  4-  1  =  5,  equal  arcs,  while  2^  4-  1  =  9,  not  being  a  prime  number, 
does  not  come  under  the  rule.  The  division  into  2*  -f  1  =  17  equal 
arcs,  not  to  speak  of  greater  numbers,  involves  a  process  too  intricate 
to  come  within  the  scope  of  an  elementary  work. 

411.  Stellate  Polygons.  If  a  circumference  be  divided 
into  n  equal  parts,  and,  beginning  at  one  of  the  points  of 
division,  we  go  round  the  circumference  drawing  chords 
subtending  these  arcs,  m  and  m,  m  being  prime  to  n  and 
<  ^  n,  we  obtain  a  polygon  of  starlike  form  called  a  stellate 
polygon. 

We  say  m  is  to  be  <  ^  n,  so  as  to  confine  our  discussion 
to  the  minor  of  the  arcs  subtended  by  each  chord,  and  con- 
sisting respectively  of  m  and  n  —  m  of  the  equal  parts.  If 
m  were  a  factor  of  n,  say  am  =  n,  then  we  should  return 
to  the  starting  point  after  placing  a  chords,  forming  a  regu- 
lar polygon  of  a  sides.  But,  if  m  is  prime  to  n,  then  as  mn 
is  the  least  number  that  divided  by  m  gives  n  as  quotient, 
we  shall  go  round  the  circumference  m  times  before  return- 
ing to  the  starting  point,  and  in  doing  so  shall  place  n 
chords. 


216  PLANE   GEOMETRY.  — BOOK    VI. 

There  is,  accordingly,  no  stellate  polygon  of  3,  4,  or  6 
vertices,  since  each  of  these  numbers  has  no  prime  to  it 
less  than  its  half. 

Since  2  is  the  only  number  prime  to  5  and  less  than  its 
half,  there  is  but  one  stellate  pentagon ;  as  ^. 

Since' 3  is  the  only  number  prime  to  8  and  less  than  its 
half,  there  is  but  one  stellate  octagon ;  as  B. 


For  the  like  reason  there  is  but  one  stellate  decagon,  and 
one  stellate  dodecagon. 

But  since  there  are  three  numbers,  2,  4,  and  7,  that  are 
prime  to  15  and  less  than  its  half,  there  are  three  stellate 
pentadecagons,  of  which  two,  C  and  B,  are  given. 


In  like  manner,  as  there  are  seven  numbers,  8,  7,  6,  5, 
4,  3,  2,  that  are  prime  to  17  and  less  than  its  half,  there  are 
seven  stellate  polygons  having  17  vertices. 

Among  the  many  methods  that  have  been  devised  for 
obtaining  an  approximate  value  of  the  important  constant 
denoted  by  tt,  that  based  upon  the  following  proposition  is 
perhaps  the  simplest,  enabling  us  to  find,  in  succession,  the 
value  of  a  side  of  a  polygon  of  2  n,  4  n,  •  •  •  sides. 


DIVISION  OF  CIRCUMFERENCE. 


217 


Proposition  XV.     Theorem. 

412.  The  radius  and  the  value  of  the  chord  of  an 
arc  being  given,  we  can  find  the  value  of  the  chord  of 
half  that  arc. 


Given :  In  circle  ACBE,  a  radius  OC  perpendicular  at  D  to  AB, 
the  chord  of  arc'^C5; 

To  Prove:  The  value  of  the  chord  of  one  half  arc  ACB  can  be 
found. 

Since  oc  is  ±  to  AB  at  D,  (Hyp.) 

AD  =  l  AB,  and  arc  ^(7  =  ^  arc  ACB.  (172) 

Join  OA,  AC.     Then  since  AC  <  Si  quadrant,* 
Zaoc<  a  rt.  Z; 


.  AC^ oOA^-\-Oc'^  —  2  0C'OD; 

(350) 

i.e.,  ac'^2r^-2R'OD. 

Since  Z  ADO  is  a  rt.  Z, 

(Hyp.) 

Olf  ^  OA^  —  Alf  =o  7^2  _  1  AJi^ ; 

(.347) 

.-.    OD=^R''-\AB-  =  ^W^R^-AB'' 

(OD,  R,  AB,  here  denoting  numerical  measures;) 

.-.  Jc''=  2  i?2  _  2  7?  .^^Ir^-Ab\' 


.'.  ac=\2r--  ie V4  R^  -  ab\ 
If,  as  is  usually  done,  we  take  7i'  =  1,  then 

AC  =^2 -^4. 


ab': 


*  At  most  AC  =  k  of  the  circumference,  since  ACB  =,  at  most,  I  of  the 
circumference. 


218  PLANE   GEOMETRY.  — BOOK   VI. 

Pkoposition  XVI.     Problem. 

413.  To  find  an  approximate  value  of  it,  the  ratio  of 
circumference  to  diameter. 


Formula  :  Sgu  =  ^2  —  V4  —  s„, 

where  s„  denotes  the  vakie  of  a  side  of  a  regular  inscribed 
polygon  of  n  sides,  R  being  taken  =  1.  If  we  take  n  =  6, 
then  s„=l  (408). 


Computation.  Val.  of  sibe. 


Value  of 

PERIM.  =  n.S 


V2  -  V4  -  1  =  .51763809.  6.21165708. 


s,,  =V2-\/4-(.5176809)--2    =.26105238.  6.26525722. 


V2  -  V4  -  (.26105238/^  =  .13080626.  6.27870041. 


s^  ^  V2  -  \/4  -  (.  13080626)2  =  .06543817.  6.28206396. 


8^^^  =  V2  -  V4  -  (.06543817)=^  =  .03272346.  6.28290510. 


V2  -  V4  -  (.03272346)2  =  .01636228.  6.28311544. 


s,^  =  V2  -  V4  -  (.01636228)2  =  .00818126.  6.28316954. 

Taking  this  last  value  of  the  perimeter  as  an  approxima- 
tion towards  the  value  of  the  circumference  whose  radius  is 

1,  since  tt  =  —  (396),  we  obtain  tt  =  1(6.28317)  =  3.14159 

Z  R 
nearly,  a  result  correct  to  the  last  decimal. 

By  the  useless  expenditure  of  much  time  and  labor,  the 
value  of  TT  has  been  calculated  as  far  as  700  places  of  deci- 
mals. The  first  fifteen,  more  than  sufficient  for  any  useful 
purpose,  are  tt  =  3.141592653589793.  Ten  decimals  are  suffi- 
,cient  to  give  the  circumference  of  the  earth  to  the  fraction 
of  an  inch,  and  thirty  decimals  would  give  the  circumference 
of  the  whole  visible  universe  to  a  quantity  imperceptible 
with  the  most  powerful  microscope. 


DIVISION  OF  CIRCUMFERENCE. 
Proposition  XVIT.     Problem. 


219 


414.  The  diameter  being  given,  to  find  a  line  ap- 
proximately equal  to  the  circumference. 


N 


C..-' 


M 


Given :  D,  the  diameter  of  a  circle  ; 

Required:  To  find  a  straight  line  approximately  equal  to  the 
circumference  of  that  circle. 

Draw  indefinite  lines  AM,  AN,  making  any  angle. 

Upon  AM  lay  off  ^^  =  113,  and  ^(7=355,  units  of  length,  =* 
and  upon  AN  lay  off  AD  =B.  Join  BD  and  draw 
CX  II  to  BD. 

Since  AB  :  AC  =  AD  :  AX  =  113 :  355,  (274) 

AX  =  ff-|  AD  =  D  X  3.141592  .  •  •.  Q.E.F. 

The  line  thus  obtained  is  greater  than  the  circumference 
by  about  one  four  millionth  of  the  diameter,  as  the  student 
may  easily  prove  for  himself. 

415.  In  the  inscribed  equilateral  tri- 
angle ABC,  draw  AD  Jl  to  BC,  and  join 
CD.  Then  AD  bisects  arc  BDC  (172), 
and  DC  =:  R,  the  radius  (408) ; 

.-.  AC^  =  da'-dc^  =  4:R'-r'-  =  3  r'; 

.:  AC  =  R^3. 


*  Lengths  respectively  equal  to  1.13  and  3.55  units  can  be  taken  from  a 
diagonal  scale ;  or  see  Exercise  836. 


220 


PLANE   GEOMETRY.  — BOOK    VI. 


416.  In  the  inscribed  square  ADBC, 
joining  AB,  CD, 

AC^z=Ad^+0C^  =  2R^  (347) ; 

.-.  ac=bV2. 

417.  In  the  inscribed   pentagon,  the 
relation   of   the    side   to   the   radius   is 
most  readily  obtained  from  the  expres-    ^ 
sion  for  the  side  of  the  inscribed  deca- 
gon (see  420) .   .  From  this  we  obtain 


AB  =  ^R  VlO  -  2  V5. 


418.  In  the  inscribed  hexagon,  the  side  DC—R  (408),  a 
result  that  can  be  obtained  also  by  means  of  the  formula 
referred  to  in  the  next  article. 


419.  In  the  inscribed  octagon  (see  diagram  for  Art.  416) 
the  value  of-  a  side  is  most  readily  found  by  means  of  the 
formula  obtained  in  Prop.  XV. ;  that  is 


AC 


=  yJ2R^-E-^4:B'' 


"ab\ 


In  this,  putting  AE  for  AC,  and  R^2,  the  value  of  a  side  of 
the  inscribed  square,  for  AB,  we  obtain 

AE  =  V2  i?2  _  22  V4  R'  -  f¥; 


i.e.,  AE  =  V2  R'  -  R''^2  =  R-^2-  V2. 

420.    In  the  inscribed  decagon  AF  •"  E  (see  diagram  for 

Art.  417), 

since  R:  BF=BF:R  —  BF,  (409) 


BF 


—    7?2 


R  •  BF; 


(322) 


.-.    BF 


R{v5  —  1),  solving  as  a  quadratic. 


EXERCISES.  221 


In  the  formula  AC  =  yl2  r""  —  R^^  R^ —  Ab"^,  putting 
-|-ie(V5  — 1),  the  value  of  a  side  of  the  inscribed  deca- 
gon, for  AC,  and  squaring,  we  obtain 


^  R\-y'o  -iy  =  2R'-  R^4.  R"-  -  Ab\' 

whence  2^4: r'-  AB^  =  R{V5  +  1); 

whence  AB  =  ^R(10  —  2  VS) . 


EXERCISES. 

QUESTIONS. 

584.  Into  how  many  equal  parts,  up  to  15,  are  we  able  to  divide  a 
circumference  by  means  of  the  straight  line  and  the  circle  ? 

585.  What  is  the  ratio  of  the  radius  of  the  circle  circumscribed 
about  an  equilateral  triangle  to  that  of  the  inscribed  circle  ? 

586.  By  which  of  the  constructions  of  this  book  can  a  right  angle 
be  divided  into  five  equal  parts  ? 

587.  How  many  different  stellate  polygons  can  be  formed  if  we 
have  a  circumference  divided  into  7  equal  parts  ?  9  equal  parts  ? 
11  equal  parts  ?     13  equal  parts  ? 

588.  If  two  nonadjacent  sides  of  a  regular  pentagon  be  produced 
to  meet,  what  angle  will  they  contain  ? 

589.  What  is  the  value  of  the  interior  angles  of  a  regular  octagon  ? 
Of  a  regular  decagon  ?  Of  a  regular  pentadecagon  ?  What  is  the 
limit  towards  which  each  of  the  interior  angles  of  a  regular  polygon 
tends  as  the  number  of  sides  increases  indefinitely?  What  is  the 
limit  for  each  exterior  angle  ? 

590.  What  is  the  circumference  and  area  of  a  circle  whose  diameter 
is  10  inches,  supposing  tt  =  3.1410? 

591.  What  is  the  circumference  and  area  of  a  circle  whose  radius 
is  2  ft.  6  in.  ? 

692.   What  is  the  circumference  of  a  circle  whose  area  is  100  sq.  in.  ? 


222  PLANE    GEOMETRY.  — BOOK   VL 

593.  What  is  the  radius  of  a  circle  whose  area  is  6  sq.  ft.  ? 

594.  What  is  the  radius  of  a  circle  whose  circumference  is  12  in,  ? 

595.  What  is  the  radius  of  a  circle  whose  area  is  ii  times  that  of  a 
circle  with  radius  B  ? 

596.  What  is  the  area  of  the  ring  between  two  concentric  circles 
whose  radii  are  5  ft.  and  7  ft.  respectively  ? 

597.  Within  a  circle  whose  radius  is  i?,  a  circle  is  drawn  so  as  to 
cover  I  of  the  surface  of  the  first  circle.  What  is  the  radius  of  the 
second  circle  ? 

598.  The  radii  of  two  similar  segments  are  as  4  to  7 ;  if  the  first 
segment  contains  25  sq.  in. ,  what  does  the  other  contain  ? 

599.  In  a  white  circle  of  3  in,  radius,  an  inscribed  square  is  painted 
black.     How  much  white  surface  will  remain  ? 

600.  In  a  white  square  whose  side  is  4  in.,  an  inscribed  circle  is 
painted  red.     How  much  white  surface  will  remain  ? 

601.  If  the  radius  of  a  circle  is  10  in.,  what  is  the  side  of  the 
inscribed  equilateral  triangle  ? 

602.  If  the  side  of  an  inscribed  equilateral  triangle  is  10  in,,  what 
is  the  radius  of  the  circle  ? 

603.  If  the  radius  of  a  circle  is  8  in.,  what  is  the  side  of  a  regular 
inscribed  pentagon  ? 

604.  If  the  side  of  a  regular  inscribed  pentagon  is  9  in.,  what  is 
the  radius  of  the  circle  ? 

605.  If  the  radius  of  a  circle  is  6  in. ,  what  is  the  side  of  a  regular 
inscribed  octagon  ? 

606.  What  must  be  the  radius  of  a  circle  so  that  a  side  of  a  regular 
inscribed  octagon  shall  be  10  in.  ? 

607.  If  the  radius  of  a  circle  is  10  in.,  what  is  the  side  of  a  regular 
inscribed  decagon  ? 

608.  What  must  be  the  radius  of  a  circle  so  that  a  side  of  a  regular 
inscribed  decagon  shall  be  3  in.  ? 

THEOREMS. 

609.  An  angle  of  a  regular  polygon  of  n  sides  is  to  an  angle  of  a 
regular  polygon  of  w  4-  2  sides,  as  n^  —  4  is  to  n^. 

610.  If  the  bisectors  of  all  the  angles  of  a  polygon  meet  in  a  point, 
a  circle  can  be  inscribed  in  that  polygon. 


EXERCISES.  223 

611.  An  inscribed  equilateral  triangle  is  equivalent  to  half  the 
regular  hexagon  of  the  same  radius. 

612.  The  altitude  of  an  equilateral  triangle  is  equal  to  the  side  of 
an  equilateral  triangle  inscribed  in  a  circle  whose  diameter  is  the  base 
of  the  first. 

613.  The  altitude  of  an  equilateral  triangle  is  to  the  radius  of  the 
circumscribing  circle  as  3  is  to  2. 

614.  The  area  of  the  regular  hexagon  is  a  mean  proportional  be- 
tween the  areas  of  the  inscribed  and  circumscribed  equilateral  triangles. 

615.  The  square  of  a  side  of  an  inscribed  equilateral  triangle  is 
equivalent  to  three  times  the  square  of  a  regular  hexagon  inscribed  in 
the  same  circle. 

616.  If  the  arcs  subtended  by  two  sides  of  an  equilateral  triangle 
be  bisected,  the  chord  joining  those  points  will  be  trisected  by  those 
sides. 

617.  The  diagonals  drawn  from  a  vertex  of  a  regular  pentagon  to 
the  opposite  vertices,  trisect  that  angle. 

618.  The  diagonals  drawn  from  a  vertex  of  any  regular  polygon  of 
n  sides  to  the  opposite  vertices,  divide  the  angle  into  n  —  2  equal 
parts. 

619.  The  diagonals  joining  alternate  vertices  of  a  regular  pentagon 
form  by  their  intercepts  a  regular  pentagon. 

620.  If  the  alternate  sides  of  a  regular  pentagon  be  produced  to 
meet,  the  points  of  meeting  will  be  the  vertices  of  another  regular 
pentagon. 

621.  The  intersecting  diagonals  of  a  regular  pentagon  divide  each 
other  in  extreme  and  mean  ratio. 

622.  In  a  regular  pentagon  ABODE,  diagonals  AC,  BE,  are  drawn, 
intersecting  in  F;  show  that  FD  is  a  parallelogram. 

623.  A  ribbon  is  folded  into  a  flat  knot  of  five  edges ;  show  that 
these  edges  form  a  regular  pentagon. 

624.  If  P,  H^  and  D  denote  respectively  a  side  of  a  regular  in- 
scribed pentagon,  hexagon,  and  decagon ;  then  P'^=o=H'^  4-  D^. 

625.  If  from  any  point  within  a  regular  polygon  of  n  sides,  perpen- 
diculars be  drawn  to  the  sides,  the  sum  of  these  perpendiculars  will 
be  equal  to  n  times  the  apothem. 


224  PLANE   GEOMETRY.  — BOOK   VI. 

626.  The  radius  of  an  inscribed  regular  polygon  is  a  mean  propor- 
tional between  its  apothem  and  the  radius  of  the  similar  circumscribed 
polygon. 

627.  The  area  of  a  circular  ring,  i.e.,  the  space  between  two  con- 
centric circumferences,  is  equal  to  that  of  a  circle  having  for  diame- 
ter a  chord  of  the  outer  circle  tangent  to  the  inner  circle. 

628.  If,  on  the  hypotenuse  and  the  arms  of  a  right  triangle  as  diame- 
ters, semicircles  be  described,  the  curvilinear  figures  bounded  by  the 
greater  and  the  two  lesser  semicircumferences  will  be  equivalent  to 
the  triangle. 

PROBLEMS. 

Inscribe  in  a  given  circle : 

629.  An  equilateral  triangle.  631.    A  square. 

630.  A  regular  pentagon.  632.   A  regular  octagon. 

Circumscribe  about  a  given  circle  : 

633.  An  equilateral  triangle.  635.    A  regular  pentagon. 

634.  A  regular  hexagon.  636.  ,A  regular  decagon. 

637.  Describe  a  square  about  a  given  rectangle. 

638.  Inscribe  an  equilateral  triangle  in  a  given  square,  so  as  to 
have  a  vertex  of  the  triangle  at  a  vertex  of  the  square. 

639.  Construct  an  equilateral  triangle  that  shall  be  double  the  area 
of  a  given  equilateral  triangle. 

•  640.    Construct  a  square  that  shall  be  |  of  a  given  square. 

641.  Construct  a  regular  pentagon  that  shall  be  |  of  a  given  regular 
pentagon. 

642.  Construct  a  regular  hexagon  that  shall  be  |  of  a  given  regular 
hexagon. 

643.  Describe  a  circle  equivalent  to  |  of  a  given  circle. 


PART   11.     SOLID    GEOMETEY. 
Book  Vll. 

PLANES  AND  POLYHEDRAL  ANGLES. 

<xStf*^Oo 

PLANES  AND  PERPENDICULARS. 

Thus  far  have  been  investigated  the  properties  of  figures 
confined  to  one  plane.  We  are  now  prepared  to  enter  upon 
the  properties  of  figures  in  space ;  hence  the  name,  Geome- 
try of  Space,  often  applied  to  this  department  of  the 
subject,  also  called  Solid  Geometry,  and  Geometry  of  Three 
Dimensions.  It  is  to  be  remembered  that,  although  in 
our  diagrams  we  can  represent  only  a  limited  portion  of  a 
plane,  the  plane  thus  represented  is  to  be  regarded  as  hav- 
ing indefinite  extension. 

Proposition  I.     Theorem. 

421.  Through  any  given  straight  line  an  infinite 
number  of  planes  can  he  passed. 


N 

For  if  we  take  any  two  points  a,  b,  in  a  plane  MN,  the 
Geom.  — 15  225 


226  SOLID   GEOMETRY.  — BOOK   VII. 

straight  line  drawn  joining  those  points  will  be  wholly  in 
that  plane  (9).  By  making  the  line  ah,  thus  drawn,  to 
coincide  with  any  given  line  AB,  we  have  one  plane  MN 
passing  through  AB.  By  revolving  MN  round  AB  as  an 
axis,  MN  can  be  made  to  occupy  any  number  of  positions, 
each  of  which  is  the  position  of  a  plane  passing  through  AB. 

Scholium.  Hence  a  plane  is  not  determined  —  that  is, 
marked  off  from  other  planes  —  by  the  single  condition  that 
it  passes  through  a  given  straight  line.  To  determine  a 
plane,  some  other  condition  must  be  given. 


Proposition  II.     Theorem. 

422.  A  plane  is  determined  hy  a  straight  line  and 
a  point  without  that  line. 


N 

Giveyi :  A  straight  line  AB,  and  a  point  C  not  in  AB  ; 
To  Prove:  Only   one   plane   can  pass  through   AB    and  also 
through  C. 

A  plane  MN  being  passed  through,  and  revolved  about, 
AB  (421)  will  have  a  determined  position  when  it  comes  to 
contain  the  point  C.  For  if  it  be  then  turned  in  either 
direction  about  AB,  it  will  cease  to  contain  the  point  C.    q.e.d. 

423.  Cor.  1.  A  plane  is  determined  hy  two  straight  lines, 
intersecting  or  parallel. 

For  it  will  be  determined  by  either  of  those  lines  and  any 
point  without  it  in  the  other  line  (422). 


PLANES  AND  PERPENDICULARS. 


227 


424.  Cor.  2.  A  plane  is  determined  by  any  three  points 
not  in  the  same  straight  line. 

For  any  two  of  the  points  may  be  joined  by  a  straight 
line,  which,  with  the  third  point,  determines  the  plane  (422). 

425.  CoR.  3.  Through  a  given  point  in  space  can  pass 
only  one  2^ai^(^llel  to  a  given  straight  line. 

Eor  in   the  one  plane  determined   by  the  line   and  the 
point,  there  can  pass  through  that  point  only  one  parallel 
^to  the  given  line  (105). 

^426.   CoR.  4.   The  intersection  of  two  planes  is  a  straight 
line. 

For  their  intersection  cannot  con- 
tain three  points  that  are  not  in  the 
same  straight  line,  seeing  that  one 
plane  and  only  one  can  pass  through 
three  such  points  (424) . 

427.  Definitions.  A  straight  line  is 
perpendicular  to  a  plane,  if  it  is  perpen- 
dicular to  every  straight  line  drawn  in  the  plane  through 
its  foot,  the  point  in  which  it  meets  the  plane.  The  plane, 
in  this  case,  is  also  said  to  be  perpendicular  to  the  line. 

Exercise  644.  In  the  diagram  for  Prop.  II.,  if  a  plane  were  passed 
through  AC,  and  turned  about  that  line  as  an  axis,  when  would  the 
revolving  plane  coincide  with  plane  MN  9  If  the  plane  continued  to 
revolve,  when  would  it  again  coincide  with  31 N  ? 

645.    The  intersection  of  a  plane  with  a  surface  on  which  no  straight 
line  can  be  drawn,  e.g.  the  surface  of  an  eggshell 
or  of  an  apple,  is  what  sort  of  a  line  ? 

646.  Enunciate  the  converse  of  Cor.  4  of  Prop. 
II.  ;  is  that  converse  generally  true  ? 

647.  If  a  circumference  be  divided  into  six  equal 
parts,  and  chords  be  drawn  subtending  these  parts, 
two  and  two,  as  in  the  accompanying  diagram,  is 
the  six-pointed  figure  thus  formed  a  stellate  polygon  according  to  the 
definition  given  in  Art.  411?    If  not,  why  not? 


228  SOLID   GEOMETRY.  — BOOK    VII. 

Proposition  III.     Theorem. 

428.  A  straight  line  perpendicular  to  each  of  two 
straight  lines  at  their  intersection,  is  perpendicular 
to  the  plane  of  those  lines. 


Given:  AP  perpendicular  to  botli  BB'  and  CC'  in  plane  MN, 
at  P; 

To  Prove :     AP  is  perpendicular  to  MN  at  P. 

Through  P  draw  any  straight  line  PB  in  MN ;  and  through 
D  draw  BBC  so  that  BB  —  BC  (219).  Join  A  with  B,  B, 
and  c. 


(353) 


Since  AB,  PB,  are  medians  of  A  ABC,  PBC,  resp.,    (Const.) 

AB^  +  Axf^2  Jff  +  2  BB' 

and  PB^ -^-PC^  <^2pb^  ^2bb'; 

.'.  ab'^-pb^-\-ac^-pc^^2ab''-2pb\  (Ax.  3) 

But  APB  and  ^P(7  are  rt  A  ;  (Hyp.) 


-9 


AB^  —  pb'^  =o  ap"^,  and  AC^  —  PC^  ^  Ap\-       (347) 


.-.  2Zp^=o=2l^'-2PiJ^- 
.-.   AP^  +  PB^  <>  AB^ ; 

.:  AP  is  ±  to  PB  ; 

.'.  AP  is  ±  to  itfiV, 


(Ax.  2,  Ax.  7) 

(352) 

Q.E.D.      (427) 


(being  ±  to  any  line  in  MN  through  P) 


PLANES  AND  PERPENDICULARS.  229 

Proposition   IV.     Theorem. 

429.  A  perpendicular  to  a  plane    can   he  drawn 
from  any  given  point  without  or  in  that  plane. 


1°.    Given:       A  point  A  without  a  plane  MN ; 

To  Prove :  A  perpendicular  to  MN  can  be  drawn  from  A. 

In  the  plane  passed  through  A  and  any  line  BC  in  MN, 
draw  AD  ±  to  BC.  From  D  draw,  in  the  plane  MN,  DE  ± 
to  BC ;  and,  in  the  plane  of  AD,  DE,  draw  AP  _L  to  DE. 
Then  ^P  is  ±  to  MN. 

For  since  ADC,  PDC,  and  APD  are  rt.  A,   (Const.) 


AC 


AC"  o^AD"  +  DC^,\ 
and  PC"  =0=  PD^  4-  -OC^' ) 

(347) 

-  PC"  =0=  ^Z)'  —  P/)"  =o  ^P"; 

(Ax.  3,  347) 

.-.  ^C''=o=  JP"  +  PC'^• 

(Ax.  2) 

.^  AP  is  _L  to  PC; 

(352) 

.-.   AP  is  ±  to  MN, 

Q.E.D.      (428) 

(being  ±  to  both  PD  and  PC.) 

2°.  Given:  A  point  P  in  plane  ifiv; 

To  Prove :  A  perpendicular  to  MN  can  be  drawn  at  P. 

From  P  draw  PD  _L  to  PC,  any  line  in  MN ;  and  in  any 
plane  intersecting  MN  in  BC,  draw  Z>.4  _L  to  BC.     Then  in 


230  SOLID   GEOMETRY.  — BOOK   VIL 

the  plane  of  PD,  DA,  draw  PA  _L  to  PD.    PA  is  ±  to  MN, 
as  may  be  proved  in  the  same  way  as 
in  1°. 

430.  CoR.  1.    There  can  be  drawn  hut           ¥- — 
one  perpendicular  to  a  given  playie  from        / 
a  point  without  the  plane.  /_ Z 

iV 

For  if  there  could  be  two  perpen- 
diculars, AP,  AQ,  from  A  to  MN,  draw  PQ.     Then,  in  the 
plane  APQ,  there  would  be  two  perpendiculars  from  A  to 
the  same  line  PQ,  which  is  impossi- 
ble (51). 

M 

431.  CoR.  2.    At  a  given  point  in  a  / 
plane  there  can  he  draivn  hut  one  per-       /     ^ 
pendicidar  to  that  plane.  N 

For  if  there  could  be  two  perpendiculars,  PA,  PB,  from 
P  in  MN,  suppose  a  plane  to  pass  through  PA,  PB,  inter- 
secting MN  in  PQ.  Then  from  the  same  point  in  PQ,  and 
in  the  same  plane,  there  would  be  two  perpendiculars  to 
PQ,  which  is  impossible  (41). 

432.  Cor.  3.  The  perpendicidar  is  the  shortest  line  that 
can  he  draimi  to  a  plane  from  a  given  p)oint  ivithout  the  plane. 

For  AP  is  shorter  than  any  oblique  line,  A  Q,  drawn  from 
A  to  PQ,  in  the  same  plane  APQ  (93). 

433.  Definition.  The  distance  of  a  point  from  a  plane  is 
the  length  of  the  perpendicular  from  the  point  to  the  plane. 


Exercise  648.  In  the  diagram  for  Prop.  III.,  the  planes  determined 
by  AB  and  P,  AC  and  P,  AD  and  P,  intersect  in  what  line  ? 

649.  In  those  planes,  if  circumferences  be  described  on  AB,  AC, 
and  AD,  as  diameters,  they  will  pass  through  what  point  and  have 
what  common  chord  ? 

650.  In  the  diagram  for  Prop.  V.,  if  AB  and  AD  be  joined,  and 
PD  =  AB,  show  that  {AD  +  AP) (AD-  AP)  =  35^ 


PLANES  AND  PERPENDICULARS.  231 

Proposition  V.     Theorem. 

434.  All  perpendiculars  drawn  to  a  given  straight 
line  at  a  given  point,  lie  in  a  plane  perpendicular 
to  the  line. 


Given:  PB,  PC,  PD,  any  perpendiculars  to  ^P  at  P; 

To  Prove :  PB,  PC,  PD,  all  lie  in  a  plane  that  is  ±  to  ^P  at  P. 

Through  PB,  PC,  pass  the  plane  3IN. 

Since  AP  is  ±  to  PB  and  PC,  (Hyp.) 

AP  is  ±  to  3IN,  and  3fN  is  ±  to  AP.   (428,  427) 

Through  ^P,  PD,  suppose  aj^lane  passed  cutting  3/iV^  in  PD'. 

Since  ^P  is  ±  to  3IM, 

AP  is  ±  to  PIJ',  a  line  in  3fN.  (427) 

But  AP  is  ±  to  PB  ;  (Hyp.) 

.♦.  PD  must  coincide  with  PD'  and  lie  in  3IN,    q.e.d. 

(since,  in  plane  AP,  PD,  there  can  be  but  one  _L  to  AP  at  P.) 

(41) 

435.  Cor.  1.  If  an  indefinite  line  PB  he  made  to  revolve  so 
as  to  remain  always  perpendicular  to  AP  at  P,  it  ivill  generate 
the  plane  MN  at  right  angles  to  AP. 

436.  Cor.  2.  Through  a  given  point  in  or  without  a  straight 
line  can  be  passed  but  one  plane  perpendicular  to  the  line. 


232  SOLID   GEOMETRY.  — BOOK    VII. 

Proposition  VI.     Theorem. 

437.  Obliques  drawn  from  a  point  so  as  to  meet  a 
plane  at  equal  distances  from  the  perpendicular,  are 
equal ;  of  obliques  meeti?i£  the  plane  at  unequal  dis- 
tances from  the  perpendicular,  the  more  remote  is 
the  greater. 


Given:  AB,  AC,  AD,  obliques  drawn  from  A  to  MN,  so  that  B 
and  C  are  equally  distant  from  the  perpendicular  AP,  but  PB  is 
greater  than  PB  ; 

To  Prove :   AB  is  equal  to  AC,  but  AB  is  less  than  AD. 

On  PD  lay  off  PE  =  PB,  and  join  AE. 

Since  PB  =  PC  =  PE,  AP  is  common,  ) 

,    ,       /  X  c  (Hyp.  and  Const.) 

and  the  z§  at  P  are  rt.  A,  ) 

AB  =  AC  =  AE.  (66,70) 

But  PE  <  PD  ;  (Const. ) 

.■.AE<AD;  (99) 

.-.  AB  <  AD.  Q.E.D. 

438.  Cor.  1.  Conversely,  equal  obliques  drawn  from  a 
point  to  a  plane  meet  it  at  equal  distances  from  the  perpendic- 
ular ;  and,  of  unequal  obliques  from  the  same  point,  the  greater 
meets  the  plane  at  a  greater  distance  from  the  perpendicular 
than  the  less. 

439.  Cor.  2.  The  locus  of  all  the  points  in  a  plane  that  are 
equally  distant  from  a  given  point  without  the  plane,  is  a  cir- 
cumference whose  center  is  the  foot  of  the  perpendicular  from 
the  given  point. 


PLANES  AND  PERPENDICULARS.  233 

440.  Cor.  3.  The  locus  of  all  the  points  in  space  that  are 
equally  distant  from  two  given  points,  is  the  plane  that  is  per- 
pendicular at  its  mid  point  to  the  straight  line  joining  the 
given  points. 


Proposition  VII.     Theorem. 

441.  //  from  the  foot  of  a  perpendicular  to  a  plane 
a  line  he  drawn  at  right  angles  to  any  line  in  the 
plane,  and  the  foot  of  this  he  joined  with  any  point 
in  the  first  perpendicular,  this  line  will  he  perpendic- 
ular to  the  line  of  the  plane. 


N 

Given:  AP  perpendicular  to  plane  MN ;  PD  perpendicular  to 
BC,  any  line  in  MN,  at  D ;  and  AD  joining  any  point  in  ^P 
with  D  ; 

To  Prove :  AD  is  perpendicular  to  BC. 

Lay  off  DB  =  DC,  and  join  A  and  P  each  with  B  and  C. 
Since  DB  =  DC  (Const.),  PD  is  common, 

and  it.  Z  PDB  =  rt.  Z  PDC,  (HyP-) 

PB=PC;  (66,70) 

.'.  AB  =  AC;  (437) 

.-.  AD  is  ±  to  BC,  Q.E.D.       (75) 

(A  and  7)  being  points  equally  distant  from  B  and  C.) 

442.  Cor.  If  from  a  point  without  a  plane  two  perpendicu- 
lars he  drawn,  one  to  the  plane  and  the  other  to  any  line  in 
the  plane,  the  line  joining  the  feet  of  these  perpendicidars 
will  he  perpendicular  to  the  line  in  the  plane;  i.e.,  if  AP  is 
J.  to  MN  and  AD  is  ±to  BC,  then  PD  is  ±  to  BC. 

.       /  OF  THE 


234 


SOLID   GEOMETRY.  — BOOK    VII. 


PLANES  AND  PARALLELS. 
Proposition  VIII.     Theorem. 

443.  Two  perpendiculars  to  the  same  plane  are  par- 
allel. 


Given:  AP  and  BQ,  each  perpendicular  to  plane  MN; 
To  Prove :  AP  is  parallel  to  B  Q. 

Through  AP,  BQ,  pass  a  plane  APB  meeting  MN  in  PQ. 
Since  AP  and  BQ  are  each  ±  to  MN,         (Hyp.) 
AP  and  BQ  are  each  ±  to  PQ  in  plane  APB  ;    (427) 
.-.   AP  is  II  to  BQ.  Q.E.D.      (106) 

444.  Cor.   1.     If  AP,  one  of  two   iiarallels,   is  A^  to  a 
playie  MN,  then  B  Q,  the  other,  is  also  _L 
to  MN.  M 

For  if  we  draw  BR  _L  to  MN  (429), 
since  BQ  is  II  to  AP  (Hyp.),  and  BR 
is    II  to  AP    (443),  BQ  must  coincide    /  -^        :r  q 

with  BR  and  be  A.  to  3IN,  since  other-  -^ 

wise  there  would  be  two  lis  to  AP  through  the  same  point 
B,  which  is  impossible  (425). 

445.  Cor.  2.   If  two  parallels,  A  and  B,  are  each  parallel  to 
a  third  line  C,  in  another  plane,  they  are  parallel  to  each  other. 

Por  if  we  pass  a  plane  J_  to  C,  it  will  also  be  J_  to  ^  and 
B  (444)  ;  whence  A  and  B  are  parallel  (443). 


PLANES  AND  PARALLELS.  235 

446.  Definition.  A  straight  line  is  parallel  to  a  plane 
if  it  cannot  meet  the  plane  though  both  be  indefinitely  pro- 
duced. 

447.  Definition.  Two  planes  are  parallel  if  they  can- 
not meet  though  indefinitely  produced. 


Proposition  IX.     Theorem. 

448.  Any  straight  line  without  a  plane  is  parallel 
to  the  plane  if  parallel  to  any  line  in  it. 

A B 

M 


\J 


N 

Given :      AB  parallel  to  CD,  a  line  in  plane  MN ; 
To  Prove :         AB  is  parallel  to  plane  MN. 

Pass  a  plane  through  AB  and  CD. 

Since  AB  is  in  the  plane  AD,  if  it  could  meet  the  plane 
MN  it  would  do  so  in  the  intersection  of  AD  with  MN,  that 
is,  in  line  CD, 

But  AB  cannot  meet  CD,  since  they  are  parallel.     (Hyp.) 

Hence  AB  cannot  meet  MN,  i.e.,  AB  is  II  to  MN.   q.e.d.    (446) 

449.  Cor.  If  a  line  AB  is  parallel  to  a  plane  MN,  and  a 
plane  be  passed  through  AB  so  as  to  intersect  MN  in  CD,  then 
CD  is  parallel  to  AB. 

For  AB,  though  in  the  same  plane  with  CD,  cannot  meet 
it,  seeing  that  CD  lies  in  MN,  which  AB  cannot  meet. 


236 


SOLID   GEOMETRY.  — BOOK    VII. 


Proposition  X.     Theorem. 

450.  Planes    that  are  perpendicular   to   the    same 
straight  line  are  parallel. 


./^^^ 

/> 

./  .   /■ 

Given:  Planes  MN  and  PQ,  each  perpendicular  to  AB ; 
To  Prove:     Plane  MN  is  parallel  to  plane  PQ. 

For  if  MN  could  meet  P  Q,  then  through  any  point  com- 
mon to  both  would  be  passed  two  planes  perpendicular  to 
the  same  straight  line  AB.  But  this  is  impossible  (436). 
Hence  MN  cannot  meet  PQ.     q.e.d. 


Proposition  XI.     Theorem. 

451.  The  intersections  of  two  parallel  planes  with 
a  third  plane  are  parallel. 

A/ 


Given:  Two  parallel  planes,  MN,  PQ,  intersecting  plane  BB  in 
AB,  DC; 

To  Prove :  AB  h  parallel  to  DC. 

Since  plane  MN  cannot  meet  plane  PQ,        (Hyp.) 
AB  cannot  meet  DC,  though  in  the  same  plane  BD  ; 

.'.  AB  is   II  to  DC.  Q.E.D.       (102) 


PLANES  AND  PARALLELS.  237 

452.  Cor.  Parallel  lines  that  are  intercepted  between  paral- 
lel planes  are  equal. 

For  since  the  plane  of  the  parallels  AD,  BC,  intersects 
the  parallel  planes  3IN,  FQ,  in  parallel  lines  AB,  DC,  the 
figure  ^C  is  a  parallelogram  (131);  whence  AD  =  BC 
(136), 


Proposition  XII.     Theorem. 

453.  A   straight  line  perpendicular  to  one  of  two 
parallel  planes  is  perpendicular  to  the  other  also. 


M' 


7f 


B  '■■•■ 


7 


Given:  Two  parallel  planes,  MN,  PQ,  and  a  straight  line  AB 
perpendicular  to  MN  ; 

To  Prove:  AB  is  perpendicular  to  PQ. 

Through  AB  pass  any  plane  ad,  intersecting  MN  in  AC, 
and  PQ  in  BD. 

Since  plane  MN  is  II  to  plane  PQ,  (Hyp.) 

ACis  II  to  bd;  (451) 

.'.  AB  is  ±  to  AC  Sind  BD;  (Hyp.,  107) 

.-.  AB  is  ±  to  plane  PQ,        q.e.d.     (427) 

(being  ±  to  any  line  through  B  in  PQ.)  ' 

454.  CoR.  Tliroxigh  a  giuen  point,  A,  one  plane  can  be 
passed  parallel  to  a  given  plane,  PQ,  and  only  one. 

For  from  A,  a  perpendicular  AB  can  be  drawn  to  PQ 
(429) ;  and  through  A,  a  plane  can  be  passed  ±  to  AB,  and 
hence  II  to  plane  PQ  (450).  Moreover,  since  from^  but  one 
perpendicular  can  be  drawn  to  PQ  (430),  there  can  be  but 
one  plane  passed  through  A  II  to  PQ. 


238 


SOLID   GEOMETRY.— BOOK   VI I. 


Proposition  XIII.     Theorem. 

455.  If  two  angles  not  in  the  same  plane  have  their 
sides  respectively  parallel  and  drawn  in  the  same 
direction,  they  are  equal. 


Given :  Two  angles,  BAC,  b'a'c',  lying  in  planes  MN  and  PQ, 
respectively,  so  that  BA  and  b'A',  ca  and  &A',  are  respectively 
parallel  and  drawn  in  the  same  direction ; 

To  Prove :    Angle  BAC  is  equal  to  angle  b'a'c'. 

In  AB,  AC,  take  any  points  B  and  C,  and  lay  oE  A'b'=AB, 
A'C'=AC;  join  AA',  BB',  CC'. 

Since  AB,  AC,  are  resp.  II  and  =  to  A'b',  a'c', 

(Hyp.  and  Const.) 

AB'  and  AC'  are  parallelograms  ;  (142) 

BB',  CC',  are  each  II  and  =  to  AA' ;  (^36) 

.-.  BB'  is  II  and  =  to  CC' ;         (445,  Ax.  1) 

.-.  BC  is  II  and  =  to  b'c' ; 

.:  ABAC  =  A  B'A'C';  (69) 

.-.  Z  A  =  Za'.  q.e.d.     (70) 

456.  Cor.  1.  If  two  angles  lying  in  different  planes  have 
their  sides  respectively  parallel,  their  planes  are  parallel. 

For  the  intersecting  lines  that  determine  the  one  plane, 
being  parallel  to  the  intersecting  lines  that  determine  the 
other,  the  planes  are  parallel. 

457.  Cor.  2.  Iftivo  parallel  planes,  MN  and  PQ,  are  inter- 
sected by  two  other  planes,  AB',  AC',  the  angles  A,  A',  formed 
by  their  intersections,  are  equal. 


PLANES  AND  PARALLELS.  239 

458.  Cor.  3.  If  three  lines,  A  A',  BB',  CC',  not  in  one 
plane,  are  equal  and  parallel,  the  triangles  ABC,  A'b'c', 
formed  by  joining  their  extremities,  are  equal,  and  their 
planes  are  parallel. 


Proposition  XIV.     Theorem. 

459.  If  two  straight  lines  are  cut  hy  three  parallel 
planes,  the  intercepts  are  proportional. 


M^ 


eI )::-3-^^ 


tN 


eI± 


Given:  A  line  AB  meeting  parallel  planes  MN,  PQ,  RS,  in  A, 
E,  B,  respectively ;  and  a  line  CD  meeting  the  same  planes  in 
C,  G,  D,  respectively ; 

To  Prove :  AE  :  EB  z=  CG  :  GD. 

Draw  AD,  meeting  PQ  in  7^;  join  AC,  EF,  FG,  and  BD. 
Since  planes  PQ,  RS,  are  II,  and  plane  ABD  cuts  them, 

EF  is  11  to  BD  ;  (451) 

.'.   AEiEB  =  AF'.FD.  (274) 

Since  planes  PQ,  MN,  are  II,  and  plane  DAC  cuts  them, 

FG  is  II  to  ^C;  (451) 

.-.    CG  :  GD  =  AF  :  FD  ;  (274) 

.-.   AE'.EB  =  CG  :  GD.  (232'") 

460.  Cor.  If  n  straight  lines  are  cut  by  m  parallel  planes, 
the  intercepts  are  jjroportional. 


240  SOLID   GEOMETRY.  — BOOK    VIL 

DIHEDRAL  ANGLES. 

461.  A  dihedral  angle  is  the  opening  between  two  planes 
that  meet.  The  line  in  which  the  planes  meet  is  called  the 
edge  of  the  angle,  and  the  two  planes 

are  called  its  faces.     Thus   the  faces  A 

AC,  BD,  meeting  in  the  edge  AB,  con-  X    \ 

tain  the  dihedral  angle  DABC.  /       A 

To  designate  a  dihedral  angle,  four  ^  V y  \  \ 

letters  are  generally  necessary,  two  at     VX      /_ \ 

the  edge  and  one  on  each  face,  the  two      \       /  \ 

at  the  edge  being  placed  between  the     b\/— ^ 

other  two.    If  the  edge  belongs  to  only 

one  angle,  the  letters  at  the  edge  will  suffice  to  designate 
the  angle.  Thus  the  dihedral  angle  DABC  may  be  referred 
to  as  dihedral  angle  AB,  or  simply  as  the  dihedral  AB. 

462.  The  plane  angle  of  a  dihedral  angle  is  the  angle  con- 
tained by  the  two  perpendiculars  drawn,  one  in  each  face,  to 
any  point  in  the  edge.  Thus  hac  is  the  plane  angle  of  the 
dihedral  DABC.  It  is  evident  that  the  plane  angle  is  the 
same  at  whatever  point  of  the  edge  it  is  constructed  (455). 

A  dihedral  angle  may  be  conceived  as  gerierated  by  a  plane 
BD  turning  from  coincidence  with  plane  AC  about  the  edge 
AB  as  axis,  till  it  reaches  the  position  where  its  plane  angle 
is  Z  bac;  which,  again,  may  be  conceived  as  generated  by  the 
revolution  of  the  line  ab  from  an  initial  position,  ac. 

463.  Two  dihedral  angles  are  equal  when  they  can  be 
placed  so  that  their  faces  coincide. 

464.  A  right  dihedral  angle  has  its  plane  angle  a  right  angle, 
and  its  faces  are  said  to  be  perpendicular  to  each  other.  In 
the  same  way,  dihedral  angles  are  acute  or  obtuse,  and  pairs 
of  dihedral  angles  are  adjacent,  complementary,  supplemen- 
tary, alternate,  corresponding,  vertical,  etc.,  according  as  their 
plane  angles  are  acute,  etc. 


.DIHEDRAL  ANGLES. 


241 


Proposition  XV.     Theorem. 

465.  Two  dihedral  angles  are  equal  if  their  plane 
angles  are  equal. 


Given:  Two  dihedral  angles,  CABD,  C'a^b^d\  having  equal 
plane  angles,  CAD,  cWd' ; 

To  Prove:  Dihedral  angle  c^ 5 Z)  =  dihedral  angle  c'a'b'd\ 

Apply  c'a'b'ij'  to  CABD  so  that  Z  c'A'd'  =^Z  CAD. 
Then  the  planes  of  these  angles  will  coincide  (423),  and 
A'b',  AB,  will  coincide,  both   being   perpendicular   to   the 
same  plane  at  the  same  point  (431)  ;  hence  the  planes  B'c' 
and  BC,  b'd'  and  BD,  will  coincide  (423) ; 

.-.  dihed.  Z  CABD  =  dihed.  Z  c'a'b'd'.     q.e.d.   (463) 


Proposition  XVI.     Theorem. 

466.  Dihedral  angles  are  to  each  other  as  their  plane 
angles. 

D 


A  D 

Given :  Two  dihedral  angles,  CABD,  c'a'b'd',  and  their  respec- 
tive plane  angles,  CAD,  c'a'd' ; 

To  Prove:  Dihedral  angle  CABD :  dihedral  angle  cU'i?'Z)'= angle 
CAD  :  angle  C'a'd'. 

Geom.  — 16 


242  SOLID   GEOMETRY.  — BOOK    VIL 

1°.  When  A  CAD  and  C'A'd'  are  commensurable. 

Let  Z  CAE  be  a  common  measure  of  these  angles,  so  that 
Z  CAE  is  contained  5  times  in  CAD  and  4  times  in  c'a'd'. 

Draw  lines  AE,  A'e',  etc.,  dividing  A  CAD,  C'A'd',  into  5 
and  4  equal  parts  respectively,  and  through  these  lines  and 
the  edges  AB,  a'b',  pass  planes  so  as  to  divide  the  dihedral 
A  CABD,  c'a'b'd',  into  5  and  4  equal  dihedral  A  respectively. 

Since  Z  CAD  :  Z  C'A'd'  =  5:  4,  (Hyp.) 

and  dihed.  Z  CABD  :  dihed.  Z  c'a'b'd'  =  5:4,    (Const.) 

dihed.  Z  CABD :  dihed.  Z  a'^'i?'//  =  Z  aiT) :  C'.l'i)'.   (232'") 

2°.  When  Z  C^D  and  C'a'd'  are    incommensurable,   we 
can   prove   by  the   method  of  limits,   as    in   (260"),   that 
always 
dihed.  Z  CABD  :  dihed.  Z  C'a'b'd'  =  Z  CAD  :  Z  cU'i)'.  q.e.d. 

467.  Cor.    Vertical  dihedral  angles  are  equal. 

For  they  have  the  ratio  of  their  equal  vertical  plane 
angles. 

468.  Scholium.  In  like  manner  may  be  established  the 
following  properties  of  dihedral  angles  by  means  of  the  cor- 
responding properties  of  plane  angles. 

(1)  All  right  dihedral  angles  are  equal. 

(2)  Ti(jo  adjacent  dihedral  angles,  formed  by  one  plane  meet- 
ing another,  are  supplementary ;  i.e.,  are  equal  to  ttvo  right 
dihedral  angles. 

(3)  Of  the  dihedral  angles  formed  by  a  plane  intersecting 
parallel  planes,  the  alternate  and  corresponding  angles  are 
equal,  and  the  interior  angles  on  the  same  side  of  the  trans- 
verse plane  are  supplementary. 

(4)  Dihedral  angles  having  their  faces  mutually  parallel, 
or,  if  their  edges  are  parallel,  respectively  perpendicular  to 
each  other,  are  either  equal  or  supplementary. 


DIHEDRAL  ANGLES, 


243 


Proposition  XVII.     Theorem. 

469.  If  a  straight  line  is  perpendicular  to  a  given 
plane,  every  plane  passed  through  that  line  is  per- 
pendicular to  the  given  plane. 


Given :  AP  perpendicular  to  plane  MN,  and  J5D,  a  plane  pass- 
ing through  AP  ; 

To  Prove :  Plane  BD  i^  perpendicular  to  plane  MN. 

Draw  PE  ±  to  5(7,  the  intersection  of  BD  and  MN. 

Since  AP  is  _L  to  plane  MN,  (Hyp.) 

^P  is  ±  to  5C  and  PE  ;  (427) 

.-.  APE,  a  right  angle, 

is  the  plane  angle  that  measures  the  dihedral  angle  formed 

by  the  planes  intersecting  in  BC ;  (462) 

.*.  plane  BD  is  ±  to  plane  MN.  (464) 

470.    Cor.   A  plane  perpendicular  to  the  edge  of  a  dihedral 
angle  is  perpendicular  to  each  of  its  faces. 


Exercise  C51.  In  the  diagram  for  Prop.  XIII.,  if  from  any  point 
D  in  A  ABC,  DD'  be  drawn  parallel  to  AA>  to  meet  A  A' B'C  in  D\ 
and  D  be  joined  with  A  and  C,  D'  with  A'  and  O',  the  triangles  DAC 
and  D'A'C  are  equal. 

652.  If  planes  are  passed  through  the  sides  of  a  triangle  perpen- 
dicular to  its  plane,  of  the  inner  dihedral  angles  formed,  no  two  are 
equal,  two  are  equal,  or  all  are  equal,  according  as  the  triangle  is 
scalene,  isosceles,  or  equilateral. 


244  SOLID   GEOMETRY.  — BOOK    VIL 

Proposition  XVIII.     Theorem. 

471.  If  two  planes  are  perpendicular  to  each  other, 
any  line  in  the  one  plane  perpendicular  to  their 
intersection  is  perpendicular  to  the  other. 


Given :  In  plane  MN,  perpendicular  to  plane  BD,  EP  perpen- 
dicular to  BC\  the  intersection  of  MN  and  BD ; 
To  Prove :       EP  is  perpendicular  to  plane  BD. 

From  P  draw,  in  plane  BD,  PA  J_  to  5C. 
Since  EP,  PA,  are  each  X  to  ^C7  at  P,    (Hyp.  and  Const.) 
APE  is  the  plane  angle  that  measures  the  dihedral  angle 
contained  by  planes  BD,  MN,  intersecting  in  BC.  (462) 
But  plane  BD  is  ±  to  plane  MN;  (Hyp.) 

.-.  APE  is  a  rt.  Z ; 
.-.  EP  is  ±  to  plane  BD,         q.e.d.     (428) 
(since  it  is  J_  to  BC  and  to  PA  in  that  plane.) 

472.  Cor.  1.  If  two  planes,  BD,  MN,  are  perpendicular  to 
each  other,  a  straight  line  PE  drawn  at  any  point  P  of  their 
intersection,  so  as  to  he  perpendicular  to  one  of  the  planes,  as 
BD,  ivill  lie  in  the  other  plane  MN. 

For  in  the  plane  of  AP  and  PE,  only  one  perpendicular 
can  be  drawn  to  AP  at  P  (431). 

473.  Cor.  2.  If  two  planes  are  perpendicular  to  each  other, 
a  perpendicular  from  any  point  of  the  one  plane  to  the  other 
must  lie  in  the  first  plane. 


DIHEDRAL   ANGLES.  245 

Proposition  XIX.     Theorem. 

474.  Through  any  straight  line  not  perperidicular 
to  a  plane^  one  plane,  and  only  one,  can  be  passed 
perpendicular  to  that  plane. 


In 


Given:  AB,  any  straight  line  not  perpendicular  to  plane  MN ; 
To  Prove :  Only  one  plane  perpendicular  to  MN  can  be  passed 
through  AB. 

Through  A  draw  ^P  ±  to  plane  MN,  (429) 

and  pass  through  AB  and  AP  a  plane,  BP. 

Since  AP  is  ±  to  plane  MN,  (Const.) 

plane  BP  is  _L  to  plane  MN.  (469) 

But  any  plane  passing  through  AB  so  as  to  be  perpendic- 
ular to  MN,  must  contain  the  perpendicular  AP  ;  (472) 
.-.  there  can  be  but  one  such  plane,     q.e.d.     (423) 

Exercise  653.  In  the  diagram  for  Prop.  XIV.,  show  that  FG  :  FE 
=  ACDF:BDAF. 

654.  Ill  the  diagram  for  Prop.  XVII.,  if  in  plane  DB,  GG'  be 
drawn  parallel  to  C'jB,  and  through  GO'  planes  be  passed  intersecting 
plane  MNin  HH'  and  KI^,  show  that  HW  is  parallel  to  KK'. 

655.  Show  that  if  HH'  and  KK'  are  on  opposite  sides  of  CB,  the 
dihedral  angles  formed  by  the  plane  GH'  and  GK'  with  J/JV,  are  or 
are  not  equal  according  as  CB  is  or  is  not  equidistant  from  HH' 
and  KK'. 

656.  Show  that,  if  BC  is  equidistant  from  HH'  and  /i/i',  plane 
DB  bisects  the  dihedral  angle  HGG'K'. 

657.  If  HH'  and  KK'  are  on  the  same  side  of  CB,  show  that  the 
dihedral  angle  formed  by  GH'  with  MN  is  the  difference  of  that 
formed  by  GK'  with  MN  and  that  formed  by  GH'  with  GK'. 


246 


SOLID  GEOMETRY.  — BOOK   VII. 


Propositiox  XX.     Theorem. 

475.  If  two  intersecting  planes  are  each  perpendic- 
ular to  a  third  plane,  their  intersection  is  also  per- 
pendicular to  the  third  plane. 


Given:  AB,  the  intersection  of  two  planes,  PQ,  RS,  each  per- 
pendicular to  plane  MN  ; 

To  Prove :      AB  [b  perpendicular  to  plane  MN. 

At  B  erect  a  perpendicular  7?  .4  to  plane  MN.    (429) 
Then  BA  lies  in  eacli  of  the  planes  PQ,  BS ;     (472) 
.*.  BA  must  coincide  with  the  intersection  of  PQ,  RS ; 

.'.  AB  is  ±  to  3fN.  Q.E.D. 

476.  CoR.  1.  A  plane  perpendicular  to  each  of  ttvo  inter- 
secting  planes  is  perpendicular  to  their  intersection. 

477.  Cor.  2.  If  a  plane  is  perpendicular  to  two  planes  per- 
pendicular  to  each  other,  the  intersection  of  any  tivo  of  these 
planes  is  perpendicular  to  the  third  plane,  and  each  intersection 
is  perpendicular  to  the  other  two. 


Exercise  658.    Prove  Cor.  2  of  Prop.  XX.  by  means  of  the  diagram 
for  Prop.  XX.,  assuming  that  plane  PQ  is  perpendicular  to  plane  RS. 

659.  In  the  same  diagram,  if  Z.  SBQ  were  140°;  what  would  be  the 
ratio  of  dihedral  angle  SBAQ  to  dihedral  angle  SBAP  ? 

660.  In  the  diagram  for  Prop.  XXI.,  show  that  the  points  P,  E,  O,  P, 
are  coney clic. 


DIHEDRAL   ANGLES. 


247 


Proposition  XXI.     Theorem. 

478.  Every  point  in  the  plane  that  bisects  a  dihedral 
angle  is  equidistant  from  the  faces  of  that  angle. 


Given :  P,  a  point  in  plane  MA,  bisecting  dihedral  angle  CABD  ; 
To  Prove:     P  is  equidistant  from  ^C  and -Bi>. 

From  P  draw  PE  A_to  AC,  and  PF  ±  to  BD  ;    (429) 
through  PE  and  PF  pass  a  plane  intersecting  AC  in  OE, 
BD  in  OF,  and  therefore  AM  in  OP. 

Since  PE  is  ±  to  ^C,  and  PF  to  BD,        (Const.) 

plane  P^T^' is  ±  to  ^^;  (476) 

.-.  POE  and  POF  are  the  plane  angles  that  measure  the 
equal  dihedral  Am  ABC,  MABD  ;  (462) 

.-.  Z  POE  =  Z.  POF ; 

.'.  rt.  A  PEO  =  rt.  A  PFO  ;  (73) 

.-.    PE  =  PF.  Q.E.D. 

479.  Definition.  The  projection  of  a  x>oint  on  a  plane 
is  the  foot  of  the  perpendicular  from  the  point  to  the 
plane. 


480.   Definition.   The  projection  of  a 
line  on  a  plane  is  the  locus  of  the  pro- 
jections on  that  plane  of  all  the  points  * 
in  the  line. 


248  SOLID   GEOMETRY.  — BOOK   VII . 

Proposition  XXII.     Theorem. 

481.  The  projection  of  a  straight  line  on  a  plane  is 
a  straight  line. 


Given :        A  straight  line  AB  and  a  plane  MN ; 

To  Prove :  The  projection  oi  AB  upon  MN  is  a  straight  line, 

Through  AB  pass  a  plane  AD  ±  to  plane  MN,  which  it 
intersects  in  CD. 

Since  AD  is  ±  to  MN,  (Const.) 

AD  contains  all  the  perpendiculars  from   points  in  AB 
upon  MN  ;  (473) 

.-.  the  feet  of  all  these  perpendiculars  must  meet  MN  in  CD. 

But  CD  is  a  straight  line ;  (426) 

.*.  the  projection  of  AB  on  MN  is  a  straight  line,    q.e.d. 

Exercise  661.  In  the  diagram  for  Prop.  XXL,  how  many  degrees 
must  there  be  in  the  plane  angle  of  dihedral  angle  CABD,  that  FO 
may  be  equal  to  FP  f 

662.  In  the  same  diagram,  how  many  degrees  are  there  in  that 
j)lane  angle  if  FF  is  equal  to  the  line  joining  FE  ? 

663.  Prove  that  if  a  line  is  equal  to  its  projection  on  a  given  plane, 
it  is  parallel  to  the  plane. 

664.  Prove  that  parallel  lines  have  their  projections  on  the  same 
plane  in  lines  that  are  coincident  or  parallel. 

6G5.  In  the  diagram  for  Prop.  XXII.,  show  that  if  Bxi  be  produced 
to  meet  MN  in  F,  E  is  a,  point  in  DC  produced. 

666.  In  the  diagram  for  the  preceding  exercise,  if  BD :  AC  =  m:n, 
what  is  the  ratio  of  CE  to  CD  f 


DIHEDRAL  ANGLES.  249 

Proposition  XXIII.     Theorem. 

482.  The  acute  angle  formed  by  a  straight  line  with 
its  own  projection  on  a  plane^  is  the  least  angle  it 
makes  with  any  line  in  that  plane. 


Given:  Angle  ABC,  formed  hj  AB  with  its  projection  SC  on 
plane  MN ; 

To  Prove :  Angle  ABC  is  less  than  angle  ABD,  formed  by  AB 
with  any  other  line  in  MN  than  BC. 

Lay  off  BD  =  BC,  and  join  AD. 

In  A  ABC,  ABD,  AB  =  AB,  BC  =  BD,      (Const.) 

hut  AC  <  AD,  (432) 

(since  ^C  is  _L  to  plane  i/^;)  (Hyp.) 

.'.   Z  ABC  <Z  ABD.  Q.E.D.        (91) 

483.  Definition.  The  acute  angle  formed  by  a  straight 
line  with  its  own  projection  on  a  plane  is  called  the  inclina- 
tion of  the  line  to  the  plane,  or  the  a7igle  of  the  line  and  plane. 

Exercise  667.  Show  that  lines  having  their  projections  on  the 
same  plane,  coincident  or  parallel,  are  not  necessarily  themselves 
parallel. 

668.  A  line  meets  a  plane  obliquely  ;  with  what  line  in  the  plane 
does  it  make  the  greatest  angle  ? 

669.  A  line  has  an  inclination  of  42°  to  each  of  two  intersecting 
planes  ;  how  many  degrees  are  there  in  the  plane  angle  of  the  dihedral 
angle  formed  by  the  planes  ? 


250  SOLID   GEOMETRY.— BOOK    VIL 

PROPOSITION  XXIV.     Theorem. 

484.  A  common  perpendicular  can  he  drawn  to  any 
two  given  straight  lines  not  in  the  same  plane. 


Given:  AB  and  (7i>,  two  straight  lines  not  in  the  same  plane  ; 
To  Prove :  A  common  perpendicular  can  be  drawn  to  AB  and  CD. 

Through  CB  pass  a  plane  MN  that  is  II  to  AB,    (454) 
and  let  ah  be  the  projection  of  AB  upon  MN. 

Since  ab  is  II  to  AB,  (449) 

ab  is  not  II  to  CD, 
(since  AB  and  CD  cannot  be  parallel ;)      (Hyp.) 
.-.  ab  will  meet  CD,  say  in  b. 
At  b  draw  bB  _L  to  ab  in  the  projecting  plane  of  AB. 
Since  AB  is  II  to  ab  in  the  same  plane, 

bB  is  ±  to  AB  ;  (107) 

and  since  bB  is  ±  to  MN,  (471) 

bB  is  ±  to  CD  also.  q.e.d.     (427) 

485.  Cor.  1.  Only  one  perpendicular  can  be  drawn  com- 
mon to  two  straight  lines  not  in  the  same  plane. 

For  if  there  could  be  two  such  perpendiculars,  then,  as 
can  easily  be  shown,  there  could  be  two  perpendiculars 
drawn  in  the  same  plane  at  the  same  point  in  a  straight 
line,  which  is  impossible  (41). 

486.  Cor.  2.  The  common  perpendicular  is  the  shortest 
distance  betiveen  two  straight  lines  not  in  the  same  plane. 


POLYHEDRAL  ANGLES.  251 

POLYHEDRAL  ANGLES. 

487.  A  polyhedral  or  solid  angle  is  the  angle  formed  by 
three  or  more  planes  meeting  in  a  common  point.  The 
point  in  which  the  planes   meet   is   called 

the  vertex;  the  intersections  of  the  planes,  ^ 

the  edges;  and  the  portions  of  the  planes  //W 

bounded  by  the  edges,  the  faces  of  the  angle.  /  /  \\ 

Thus,  in  the  polyhedral  angle  S-ABCDE,  S      V^  ■//:\\ 
is  the  vertex;  SA,  SB,  etc.,  are  the  edges;      /  \,/     \.A 
and  ASB,  BSC,  etc.,  are  faces  or  face  angles.  j       \ 

It  is  to  be  noted  that,  in  a  polyhedral  angle, 
every  two  adjacent  edges  form  a  face  angle ;  and  every  two 
adjacent  faces,  a  dihedral  angle.  It  is  also  to  be  noted  that 
the  faces  and  edges  of  a  polyhedral  angle  may  be  supposed 
to  extend  indefinitely.  As  a  convenience  in  demonstration, 
however,  portions  of  the  faces  and  edges  may  be  repre- 
sented as  cut  off  by  a  plane.  The  section  formed  ,by  the 
intersection  of  the  plane  with  the  faces  is  a  polygon, 
sometimes  called  the  base  of  the  polyhedral  angle. 

488.  A  polyhedral  angle  is  convex  if  any  section  made  by  a 
plane  cutting  all  its  faces  is  a  convex  polygon;  as  ABODE. 
It  is  to  be  understood  that  the  polyhedral  angles  about  to  be 
treated  of  are  convex. 

489.  A  polyhedral  angle  is  trihedral,  tetrahedral,  etc., 
according  as  it  has  three,  four,  etc.,  faces. 

490.  A  trihedral  angle  is  rectangular,  hirectangular,  or 
trirectangular,  according  as  it  has  one,  two,  or  three  right 
dihedral  angles.  The  ceiling  and  walls  of  a  room  form 
trirectangular  angles. 

Exercise  070.  In  the  diagram  for  Prop.  XX.,  how  many  trihedral 
angles  are  represented,  with  what  common  vertex  ? 

071.  In  the  same  diagram,  if  SBQ  is  a  right  angle,  of  what  class, 
according  to  Art.  490,  is  each  of  the  four  trihedral  angles  ? 


252  SOLID   GEOMETRY.— BOOK   VII. 

Proposition  XXV.     Theorem. 

491.  In  a  trilwdral  angle^  the  sum  of  any  two  of 
the  face  angles  is  greater  than  the  third  face  angle. 


CHven:  In  trihedral  angle  s-Abc,  face  angle  CSA  greater  than 
angle  ASB  or  angle  BSC; 

To  Prove :  Angle  ASB  +  angle  BSC  is  greater  than  angle  CSA. 

In  face  CSA  make  Z  ASB  =  Z  ASB  ;  (203) 

through  any  point  D  of  SB  draw  ABC  in  plane  CSA; 

take  SB  =  SB,  and  join  AB,  BC. 
Since  SB  =  SB,  SA  =  SA,  and  Z  ASB  =  Z  ASB,  (Const.) 
A  ASB  —  A  ASB,  and  AB  =  AB.  (66,  70) 

lliAABC,AB-{-BC>AC;  (88) 

.'.   BC>{AC  —  AB)  ov  BC.  (Ax.  5) 

In  A  BSC,  BSC,  since  SB  =  SB,  sc  =  sc, 
but  BC  >  BC, 

Zbsc>  Z  BSC;  (91) 

.-.  Z  ASB  +  Z  BSC  >  Z  ^5fz>  4-  Z  D^ic;  (Ax.  4) 

i.e.,  Z  ASB  +  ZbsoZ  CSA.  Q.E.D. 

Exercise  672.  A  plane  can  be  passed  perpendicular  to  only  one 
edge  or  to  two  faces  of  a  polyhedral  angle. 

673.  If  three  lines  in  space  are  parallel,  or  meet  in  a  common  point, 
how  many  planes  may  they  define,  taken  two  and  two  ? 


POLYHEDRAL  ANGLES.  253 

Proposition  XXVI.     Theorem. 

492.  The  sum  of  the  face  angles  of  any  convex  poly- 
hedral angle  is  less  than  four  right  angles. 

s 


Given:  ASB,  BSC,  CSD,  etc.,  face  angles  of  a  polyhedral  angle 
S-ABCDE  ; 

To  Prove :  The  sum  of  the  angles  ASB,  Bsa,  CSD,  etc.,  is  less 
than  four  right  angles. 

Pass  a  plane  so  as  to  cut  the  edges  in  A,  B,  C,  I),  E,  and 
the  faces  in  AB,  BC,  CD,  DE,  EA  ; 

then  ABODE  is  a  convex  polygon.  (Hyp.) 

Taking  any  point  0  in  ABODE,  join  OA,  OB,  00,  OD,  OE, 
As  the  number  of  triangles  having  their  vertices  at  S  is 
equal  to  the  number  of  triangles  having  their  vertices 
at  O, 
(they  having  the  same  bases  AB,  BO,  etc.,) 
the  sum  of  the  interior  A  of  the  one  set  is  equal  to  that 
of  the  other  set. 
But  in  the  trihedral  A  formed  at  A,  B,  O,  etc., 


Z  SB  A  4-  A   SBO  >  A  ABO, 
and  A  SOB  +  A  sod  >  A  bod,  etc. ; 


}  (491) 


.-.  the  sum  of  the  A  at  the  bases  of  the  triangles  whose 

vertices  are  at  S,  is  greater  than  that  of  the  triangles 

whose  vertices  are  at  0 ; 

'.  the  sum  of  the  Ant  s  <  the  sum  of  the  A  at  o.   (Ax.  5.) 

But  the  sum  of  the  AdA,  O  =  four  rt.  A^ ; 

,%  the  sum  of  the  A2X  s  <,  lour  rt.  A         q.e.d. 


254  SOLID   GEOMETRY.  — BOOK    VII. 

Proposition  XXVII.     Theorem. 

493.  //  the  edges  of  a  trihedral  angle  he  produced 
through  the  vertex,  they  will  form  the  edges  of  a 
second  trihedral  angle,  called  the  symmetrical  trihedral 
of  the  first,  with  its  face  and  dihedral  angles  respec- 
tively equal  to  those  of  the  first,  hut  arranged  in 
reverse  order. 


Given:  Trihedral  angle  s-A'b*&,  formed  by  producing  the 
edges  of  trihedral  angle  S-ABC ; 

To  Prove:  The  face  angles  and  dihedral  angles  of  S-A'b'c' 
are  respectively  equal  to  those  of  S-ABC,  but  in  reverse  order. 

1°.  The  face  A  A' SB',  ASB,  etc.,  are  respectively  equal, 

(50) 
(being  vertical  A  in  planes  determined  by  AA',  bb\  etc.) 

2°.  The  dihedral  A  SA',  SA,  SB',  SB,  etc.,  are  respectively 
equal,  (467) 

(being  vertical  dihedral  angles.) 

3°.         The  angles  of  both  kinds  are  in  reverse  order. 

For  to  meet  the  faces  of  the  first  trihedral  in  the  order 
ASB,  BSC,  OS  A,  we  must  turn  from  right  to  left,*  while  to 
meet  the  faces  of  the  second  trihedral  in  the  order  A' SB', 
B'SC',  C'SA',  we  must  turn  from  left  to  right. 

*  In  order  to  see  this,  look  at  each  trihedral  having  the  vertex  a.bove^ 


POLYHEDRAL  ANGLES.  255 

The  matter,  which  is  of  some  importance,  may  be  made  clearer  by 
the  following  illustration.  Let  x\BC,  A'B'C,  abc,  be  three  triangles 
having  their  sides  and  angles  respectively  equal ;  ABC,  A'B'C,  hav- 
ing theirs  in  the  same  order,  but  abc  in  reverse  order.  It  is  obvious 
that  ABC  can  be  made  to  slide  over  and  coincide  with  A'B'C  without 
reversing,  while  abc  must  be  taken  up  and  reversed  in  order  to  apply 
it  to  A'B'C,  either  from  above  or  below,  so  as  to  make  them  coincide. 

S  S'  s 


If,  now,  we  conceive  planes  to  be  passed  through  the  sides  of  the  tri- 
angles and  through  points  S,  S',  s,  similarly  situated  with  respect 
to  their  vertices,  it  is  evident  that  the  trihedral  angles  thus  formed  will 
have  their  face  angles  and  dihedral  angles  respectively  equal,  but  those 
of  s-abc  in  reverse  order  from  those  of  S'-A'B'C  and  8-ABC. 
Hence,  if  A  ABC  be  made  to  coincide  with  A  A'B'C,  8-ABC  will 
coincide  with  S'-A'B'C ;  since  S  will  then  coincide  with  S'.  But  if 
abc  be  made  to  coincide  with  A'B'C,  s  will  be  on  that  side  of 
A'B'C  which  is  remote  from  S',  as  in  the  figure. 

494.  Two  polyhedral  angles  are  equal  and  can  be  made  to 
coincide  if  their  face  angles  and  dihedral  angles  are  respec- 
tively equal  and  arranged  in  the  same  order  ;  if  these  parts 
are  equal  but  not  arranged  in  the  same  order,  the  poly- 
hedral s  are  said  to  be  symmetrical. 

Exercise  074.  If  four  lines  in  space  are  parallel,  or  meet  in  a 
common  point,  how  many  planes  may  they  define,  taken  two  and  two  ? 

675.  A  line  parallel  to  two  intersecting  planes  is  parallel  to  their 
intersection. 

676.  If  two  unequal  similar  triangles  not  in  the  same  plane  have 
their  sides  respectively  parallel,  the  lines  joining  their  homologous 
vertices  will,  if  produced,  meet  in  one  point. 


256  SOLID   GEOMEriiY.  —  BOOK    VII. 

Pkoposition  XXVIII.     Theorem.  , 

495.    Trihedral  angles  that  have  their  face  angles 
respectively  equal  are  equal  or  syimnetrical. 


Given:    Two  trihedral  triangles,  S-ABC,  S^-A^b'c',  having 
ASB  equal  to  A's'b',  bsc  equal  to  B's'c',  CSA  equal  to  C's'j'; 

To  Prove :  S-ABC  and  s'-a'b'c'  are  equal  or  symmetrical. 

On  the  six  edges  lay  off  SA  =  s'a',  SB  =  s'b',  SC  =  s'c\ 
and  join  AB,  AC,  BC,  a'b',  a'c',  b'c'. 

Since  A  SAB,  SBC,  SCA  =  As'a'b',  s'b'c',  s'c'a',  re- 
spectively, (66) 

AB,  AC,  BC  =  A'b',  A'c',  b'c',  respectively ; 

.-.  A  ABC  =  A  A'b'c'  Siiid  Z  B AC  =Zb'a'c'.        (69) 

At  any  point  D  in  SA  draw  1)E,  DF,  each  _L  to  SA,  in  the 
faces  ASB,  ASC  respectively.  These  lines  will  meet 
AB  and.  AC  respectively,  (114) 

(since  the  A  SAB,  SAC,  are  acute,  being  base  A  of  isos.  A.) 

Let  them  meet  AB,  AC,  in  E  and  F,  respectively,  and  join  FF. 

On  s'a'  lay  off  a'd'  =  AI),  and  construct  d'e'f'  as  DEF 
was  constructed.     Then  since 

AD  =  A'b'  (Const.)  and  Z  DAE  =  Z  d'a'e', 

rt.  AADE  =  rt.  Aa'd'e';  (63) 

.-.  AE  =  A'E'  and  BE  =  D'E'. 


POLYHEDRAL  ANGLES.  257 

In  like  manner  it  may  be  shown  that 

AF  =  .4'^'  and  DF  =  D^F\ 

Since  AE  =  A'e\  AF=A'f'.  and  Z  BAC=Z  b'a'c',    (Above) 

A  AEF  =  AA 'e'f',  and  EF  =  E'f'.  (6(j) 

Since  DE  =  J)'e',  T)F  =  d'f',  and  EF  =  e'f', 

A  DEF  =  A  d'e'f',  and  Z  i^Z»^  =  Z  F'd'e' ;       (69) 

.-.  dihedral   Z  ^S  =  dihedral  Z  ^'s',  (465) 

(being  measured  by  equal  A  FDE,  f'd'e'.) 

In  the  same  way  it  may  be  proved  that 

dihed.  Zbs  =  dihed.  Z  B's',  and  dihed.  Z  CS  =  dihed.  Z  C's'. 

Hence  the  trihedrals  S-ABC,  s'-A'b'c',  are  equal  or  sym- 
metrical according  as  the  equal  parts  are  or  are  not 
arranged  in  the  same  order.  q.e.d.     (494) 

496.  CoR.  1.  If  two  trihedral  angles  have  the  three  face 
angles  of  the  one  respectively  equal  to  the  three  face  angles  of 
the  other,  the  dihedral  angles  of  the  one  are  respectively  equal 
to  the  dihedral  angles  of  the  other. 

497.  Cor.  2.  An  isosceles  trihedral  angle,  that  is,  one  having 
two  of  its  face  angles  equal,  is  equal  to  its  symmetrical  trihedral. 

For  if  in  S-ABC,  we  have  Z  ASB  =  Z  BSC,  then  in 
s'-a'b'c'  we  shall  have  Z  a's'b' =  Z  B's'c' ;  and  the 
A  CSA,  C's'a'  will  have  equal  faces  on  each  side  of  them ; 
then  also  the  dihedral  angles  will  be  similarly  arranged,  ami 
the  trihedrals  will  be  equal  (495). 

Exercise  677.  If  a  line  makes  an  acute  angle  with  a  plane,  every 
plane  with  which  it  makes  the  same  angle  is  parallel  to  the  first. 

678.  Parallel  lines  that  intersect  the  same  plane  make  equal  angles 
with  it. 

679.  If  the  projections  of  any  number  of  points  upon  a  plane  lie  in 
one  straight  line,  these  points  are  in  one  plane.     What  plane  ? 

Geom.  — 17 


258  SOLID  GEOMETRY.  — BOOK   VII. 

EXERCISES. 

QUESTIONS. 

680.  What  space  concepts  are  determined  in  position  by  one  point, 
by  two,  by  three,  respectively  ? 

681.  To  what  theorem  in  Book  I.  does  Prop.  V.  correspond  ? 

682.  What  is  the  locus  of  all  the  points  in  space  that  are  equidis- 
tant from  a  given  circumference  ? 

683.  What  is  the  locus  of  all  the  points  in  space  that  are  equidis- 
tant from  the  vertices  of  a  triangle  ? 

684.  What  is  the  locus  of  all  the  points  in  space  that  are  equidis- 
tant from  the  mid  points  of  the  sides  of  a  triangle  ? 

685.  To  what  theorem  in  Book  I.  does  Cor.  3  of  Prop.  VI.  corre- 
spond ? 

686.  What  is  the  locus  of  all  the  lines  that  are  perpendicular  to  a 
given  line  at  a  given  point  ? 

687.  To  what  theorems  in  Book  I.  do  Prop.  VIII.  and  its  first 
corollary  correspond  ? 

688.  What  is  the  locus  of  all  the  lines  that  have  a  given  line  in  a 
given  plane  as  their  projection  ? 

689.  What  is  the  locus  of  all  the  points  in  space  that  are  at  a 
given  distance  from  a  given  plane  ?  . 

690.  To  what  theorems  in  Book  I.  do  Prop.  XII.,  Cor.,  and  Prop. 
XIII.  correspond  ? 

691.  To  what  theorem  in  Book  IV.  does  Prop.  XIV.  correspond  ? 

692.  To  what  theorem  in  Book  I.  does  Prop.  XIX.  correspond  ? 

693.  What  is  the  locus  of  all  the  points  in  space  that  are  equidis- 
tant from  two  intersecting  planes  ?    From  two  parallel  planes  ? 

THEOREMS. 

694.  If  a  line  is  perpendicular  to  one  of  two  intersecting  planes, 
its  projection  on  the  other  plane  is  perpendicular  to  the  intersection. 

695.  A  plane  parallel  to  two  sides  of  a  quadrilateral  in  space,  —  that 
is,  a  quadrilateral  having  its  sides  two  and  two  in  different  planes,  — 
divides  the  other  two  sides  proportionally. 


EXERCISES.  259 

696.  The  mid  points  of  the  sides  of  a  quadrilateral  in  space  are  the 
angular  points  of  a  parallelogram. 

697.  If  the  intersections  of  any  number  of  planes  are  parallel,  the 
perpendiculars  drawn  to  these  planes  from  the  same  point  in  space  are 
in  the  same  plane. 

698.  If  a  line  is  equally  inclined  to  both  faces  of  a  dihedral  angle, 
the  points  in  which  it  meets  the  faces  are  equally  distant  from  the 
edge  of  the  dihedral. 

699.  If  a  line  makes  equal  angles  with  three  lines  in  the  same 
plane,  it  is  perpendicular  to  that  plane. 

700.  If  a  plane  be  passed  through  one  diagonal  of  a  parallelogram, 
the  perpendiculars  to  that  plane  from  the  extremities  of  the  other 
diagonal  are  equal. 

701.  If  from  a  point  within  a  dihedral  angle,  perpendiculars  are 
drawn  to  its  faces,  the  angle  contained  by  these  perpendiculars  is 
equal  to  the  plane  angle  of  the  adjacent  dihedral  angle  formed  by  pro- 
ducing one  of  the  planes. 

702.  If  three  planes  have  a  common  intersection,  perpendiculars 
to  these  planes  from  any  point  in  the  intersection  are  in  the  same 
plane. 

703.  In  any  trihedral  angle,  the  three  planes  bisecting  its  dihedral 
angles  intersect  in  the  same  line. 

704.  In  any  trihedral  angle,  the  three  planes  passed  through  its 
edges  perpendicular  to  the  opposite  faces,  intersect  in  the  same  line. 

705.  In  any  trihedral  angle,  the  three  planes  passed  through  the 
edges  and  the  bisectors  of  the  opposite  face  angles,  intersect  in  the 
same  line. 

706.  In  any  trihedral  angle,  the  three  planes  passed  perpendicularly 
through  the  bisectors  of  the  face  angles,  intersect  in  the  same  line. 

LOCI. 

Find  the  loci  of  the  points  in  space  that  respectively  satisfy  the 
following  conditions : 

707.  Are  equidistant  from  two  given  points. 

708.  Are  equidistant  from  two  given  intersecting  lines. 

709.  Have  their  distances  from  two  given  planes  in  a  given  ratio. 


260  SOLID   GEOMETRY.  — BOOK    VI I. 

710.  Are  equidistant  from  the  vertices  of  a  given  triangle. 

711.  Are  equidistant  from  the  sides  of  a  given  triangle. 

712.  Are  equidistant  from  the  vertices  of  a  quadrilateral  whose 
opposite  angles  are  supplementary. 

713.  Are  equidistant  from  the  circumference  of  a  given  circle. 

714.  Are  equidistant  from  three  given  planes. 

715.  Are  equidistant  from  the  edges  of  a  given  trihedral  angle. 

716.  Are  equidistant  from  two  given  planes  and  two  given  points 
in  space. 

PROBLEMS. 

In  the  construction  of  the  following  problems,  it  is  assumed  that, 
besides  the  constructions  of  Plane  Geometry,  we  are  able  :  (1)  to  pass 
a  plane  through  any  given  line  and  any  point  or  line  that  can  be  in 
the  same  plane  with  it  (422,  423,  424)  ;  (2)  through  any  given  point 
in  or  without  a  given  plane,  to  draw  a  perpendicular  to  that  plane. 

717.  Through  a  given  line  in  a  plane  pass  a  plane  making  a  given 
angle  with  that  plane. 

718.  Through  a  given  line  without  a  given  plane  pass  a  plane  mak- 
ing a  given  angle  with  that  plane. 

719.  Through  the  edge  of  a  given  dihedral  angle  pass  a  plane  bisect- 
ing that  angle. 

720.  Through  a  given  point  without  a  given  plane  pass  a  plane 
parallel  to  that  plane. 

721.  At  a  given  distance  from  a  given  plane  pass  a  plane  parallel 
to  that  plane. 

722.  Through  a  given  point  pass  a  plane  perpendicular  to  a  given 
straight  line. 

723.  Through  the  vertex  of  a  given  trihedral  angle  draw  a  line  mak- 
ing equal  angles  with  the  edges. 

724.  In  a  given  plane  find  a  point  such  that  the  lines  drawn  to 
it  from  two  given  points  without  the  plane,  shall  make  equal  angles, 
with  the  plane.     (Two  cases.) 

725.  In  a  given  plane  find  a  point  equidistant  from  three  given 
points  without  the  plane. 

726.  In  a  given  straight  line  find  a  point  equidistaat  from  two 
given  points  not  in  the  same  plane  as  the  line. 


Book  VIIL 
polyhedrons. 


498.  A  polyhedron  is  a  solid  bounded  by  four  or  more  poly- 
gons. The  bounding  polygons  are  called  the  faces;  their 
intersections,  the  edges;  and  the  intersections  of  the  edges, 
the  vertices  of  the  polyhedron. 

The  least  number  of  planes  that  can  form  a  solid  angle  — 
the  trihedral  —  is  three.  As  the  faces  and  edges  extend 
indefinitely,  the  space  within  the  angle  is  of  indefinite 
extent,  so  that  in  order  to  cut  off  a  definite  portion  of  that 
space,  a  fourth  plane  must  be  passed  intersecting  the  faces. 
Hence  four  is  the  least  number  of  planes  that  can  inclose  a 
space. 

499.  A  polyhedron  of  four  faces  is  called  a  tetrahedron; 
one  of  six  faces,  a  hexahedron;  of  eight  faces,  an  octahe- 
dron; of  twelve  faces,  a  dodecahedron;  of  twenty  faces,  an 
icosahedron. 

500.  A  polyhedron  is  convex  when  every  section  of  it  by 
a  plane  is  a  convex  polygon.  As  none  but  convex  poly- 
hedrons are  to  be  treated  of  in  what  follows,  the  term  poly- 
hedron will  always  signify  convex  polyhedron. 

501.  The  volume  of  a  polyhedron  is  its  quantity  as  meas- 
ured by  the  polyhedron  taken  as  unit  of  volume,  or  is  the 
numerical  measure  of  that  quantity. 

In  every-day  life,  volume  is  expressed  by  stating  how 
many  stated  solid  measures  a  given  solid  contains;  as  356 
cu.  in.     In   abstract   discussions,   however,    by   volume   is 

261 


262  SOLID   GEOMETRY.  —  BOOK   VIII. 

meant  the  numerical  measure  of  a  solid,  or  the  ratio  of  a 
given  solid  to  the  solid  unit. 

502.  Similar  polyhedrons  \i2iYQ  fhQ  ^2imQ  form  ;  equivalent 
polyhedrons  have  the  same  volume;  equal  polyhedrons  have 
the  same  form  and  volume. 

PRISMS. 

503.  A  prism  is  a  polyhedron  two  of  whose  faces  are  equal 
polygons  having  their  homologous  sides  parallel,  and  whose 
other  faces  are  parallelograms. 

504.  The  bases  of  a  prism  are  its  equal  paral- 
lel faces  ;  the  other  faces  are  called  lateral 
faces.  The  intersections  of  the  bases  and  lat- 
eral faces  are  called  basal  edges;  the  intersec- 
tions of  the  lateral  faces  are  called  lateral  edges. 

505.  Prisms  are  triangular,  quadrangular , pentagonal,  etc., 
according  as  their  bases  are  triangles,  quadrangles,  penta- 
gons, etc. 

506.  A  right  prism  has  its  lateral  edges  perpendicular  to 
its  bases;  all  other  prisms  are  called 

oblique  prisms. 

507.  A  regular  prism  is  a  right  prism 
whose  bases  are  regular  polygons. 

CoR.     The  lateral  faces  of  a  regular 
prism  are  equal  rectangles. 

508.  The  altitude  of  a  prism  is  the  perpendicular  distance 
between  its  bases. 

Cor.  Each  lateral  edge  of  a  right  prism  is  equal  to  the 
altitude  ;  each  lateral  edge  of  an  oblique  prism  is  greater  than 
the  altitude. 

509.  A  right  section  of  a  prism  is  a  section  perpendicular 
to  its  lateral  edges. 


PRISMS. 


Proposition  I.     Theorem. 


510.  The    sections  of  a   prism    made    by  parallel 
planes  are  equal  polygons. 


Given:  A  prism  MN  intersected  by  parallel  planes  ABD,  dbd; 
To  Prove:  ^5Ci)^  is  equal  to  a6cde. 

Since  plane  ABB  is  II  to  plane  abd,  (Hyp.) 

AB,  BC,  CD,  etc.,  are  II  to  ab,  be,  cd,  etc.,  resp.,  (451) 
and  these  lines  are  similarly  directed  ; 
.-.  ZA  =  Za,  ZB=Zb,  Zc  =  Zc,  etc. ;      (455) 
.*.  ABCDE  and  abode  are  mutually  equiangular. 

Since  AB  =  ab,  BC=  be,  CD  =  cd,  etc.,  (136) 

ABCDE  and  abcde  are  mutually  equilateral; 

.-.  ABCDE  =  abcde.  q.e.d.     (61) 

511.  Cor.     Any  section  of  a  prism  parallel  to  the  base  is 
equal  to  the  base ;  also  all  right  sections  of  a  prism  are  equal. 

512.  Definition.      The  lateral  or  convex  stirface  of  a 
prism  is  the  sum  of  its  lateral  faces. 

Exercise  727.    In  the  diagram  above,  if  the  prism  MNia  pentag- 
onal, what  is  the  sum  of  the  plane  angles  of  the  lateral  dihedral  angles  ? 

728.  Supposing,  as  above,  that  MN  is  pentagonal,  how  many  faces, 
face  angles,  dihedral  and  trihedral  angles,  has  MN9 

729.  If  the  base  have  n  sides,  how  many  faces,  etc.,  has  MN ? 


264  SOLID   GEOMETRY.  — BOOK    VI IL 

Propositiox  II.     Theorem. 

513.  The  lateral  surface  of  a  prism  is  equivalent  to 
the  rectangle  contained  hy  a  lateral  edge  and  the 
perimeter  of  a  right  section  of  the  prism. 


B  c 

Given:  A  right  section  ahcde,  and  AA\  ?i,  lateral  edge  of  prism 
AD' ; 

To  Prove:  The  lateral  surface  of  AD'  is  equivalent  to  rect- 
angle A  A'  •  (ah  -\-hc  -\-  etc.). 

Since  AD'  is  a  prism,  (Hyp.) 

AB',  bc',  CD',  etc.,  are  parallelograms.  (503) 

Since  abcde  is  a  right  section,  (Hyp.) 

ab,  he,  cd,  etc.,  are  ±  to  AA',  BB',  cc',  etc.,  resp. ;  (509) 

.-.  AB'  ^  AA'  ■  ah;  BC'  =o  BB' '  bc,  etc. ; 

.-.  AB'  +  BC'  4-  CD',  etc.,  ^  AA'  -  ah  +  BB' .  bc  +  cc'  •_  cd, 

etc. ;  (Ax.  2) 

.-.  lat.  surf,  of  ad'  ^  rect.  AA'  •  (ah  +  6c  +  etc.),  q.e.d. 

(since  AA'  =  BB'  =  CC'  =  etc.)  (136) 

614.  CoR.  The  lateral  surface  of  a  right  prism  is  equivalent 
to  the  rectangle  of  its  altitude  and  the  perimeter  of  its  base. 

Scholium.  Prop.  II.  may  also  be  expressed  under  the 
form :  The  lateral  area  of  a  prism  is  equal  to  the  product 
of  a  lateral  edge  and  the  perimeter  of  a  right  section. 

In  the  same  way,  in  the  corollary  to  Prop.  II.  and  similar 
theorems,  area  may  be  substituted  for  surface,  and  p)roduct 
for  rectangle,  as  explained  in  Art.  324. 


PRISMS. 


265 


Proposition  III.     Theorem. 

515.  Two  prisms  are  equal,  if  three  faces  including 
a  trihedral  angle  of  the  one  are  respectively  equal  to 
three  similarly  arranged  faces  including  a  trihedral 
angle  of  the  other. 


Given:   In  the  prisms  Ad,  A'd',  the   faces  AD,  Ah,  Ae,  re- 
spectively equal  to  the  faces  A'd',  A'h\  A'e',  and  similarly  arranged; 

To  Prove :     Prism  Ad  is  equal  to  prism  A'd'. 


(Hyp.) 
(495) 


.     Since  AD  =  A'd',  Ah  =  A'h',  Ae  =  A'e', 

and  they  are  similarly  arranged, 

trihedral  A  A  •=  trihedral  Z  A'. 

Hence,  applying  trihed.  Z  A  to  trihed.  Z  A',  they  will 
coincide ; 

then  since  AD  ^  A'd',  Ah  ^  A'h',  and  Ae  ^  A'e', 

edge  ah  =^  edge  a'h',  and  edge  ae  =^  a'e' ; 

.'.  base  ad  ^  base  a'd', 

(being  equal  polygons  having  two  sides  coinciding ;) 

.-.  all  the  lateral  edges  will  coincide, 

(since  their  extremities  coincide;) 

.-.  the  prisms  coincide  and  are  equal,     q.e.d.  (61) 


266  SOLID   GEOMETRY.— BOOK   VIII. 

516.  Definition.   A  truncated  prism  is  a  part  of  a  prism 
cut  off  by  a  plane  not  parallel  to  the  base. 

517.  CoR.  1.  Two  truncated  prisms  are  equal, 
if  the  three  faces  including  a  trihedral  angle  of 
the  one  are  respectively  equal  to  the  three  faces 
ijicluding  a  trihedral  angle  of  the  other,  and  are 
similarly  arranged.  ^^^ 

518.  CoR.  2.   Tioo  right  prisms  are  equal  if  they  have  equal 
bases  and  equal  altitudes. 


Proposition  IV.     Theorem. 

519.  An  oblique  prism  is  equivalent  to  a  ri£ht  prism 
having  as  base  a  right  section,  and  as  altitude  a  lat- 
eral edge,  of  the  oblique  prisma. 


Given:  In  oblique  prism  AD',  a  right  section,  ahcde,  and  a  lat- 
eral edge,  AA^ ; 

To  Prove:  AD'  is  equivalent  to  a  right  prism  having  as  base 
ahcde,  and  an  altitude  equal  to  AaK 

Produce  AA'  to  a',  so  that  aa'  =  AA' ;  and  through  a'  pass 
a  plane  X  to  aa',  and  intersecting  all  the  faces  of  the  prism 
produced,  thus  forming  a  second  right  section,  a'b'c'd'e',  par- 
allel and  equal  to  ahcde. 

The  prism  ad'  thus  formed  is  a  right  prism  (506),  whose 
base  is  the  given  right  section,  ahcde,  and  whose  altitude 
aa'=^AA'  (Const.). 


PRISMS. 


267 


Now  the  figures  ABCDE-d  and  A'B'c'D'E^-d'  are  equal 
truncated  prisms  (517) ;  and,  if  to  each  of  these  we  add  the 
figure  abcde-D',  we  obtain  (Ax.  2) 

prism  ad'  <^  rt.  prism  abcde-d'.      q.e.d. 

520.  Definition.  A  parallelopiped  is 
a  prism  whose  bases  are  parallelograms. 

Cor.  Any  two  opposite  faces  of  a  paral- 
lelopiped are  equal  parallelograms. 

521.  Definition.    A  right  parallelopiped 
has  its  lateral  edges  _L  to  its  bases. 

522.  Definition.  A  rectangular  paral- 
lelopiped is  a  right  parallelopiped  whose 
bases  are  rectangles. 

Cor.   Hence  all  its  faces  are  rectangles. 


Proposition  V.     Theorem. 

523.  The  plane  passed  through  two  diagonally  op- 
posite edges  of  a  parallelopiped  divides  it  into  two 
equivalent  triangular  prisms. 


Given:  A  plane  AC'  through  diagonally  opposite  edges  of  paral- 
lelopiped bd'  ; 

To  Prove:  Triangular  prism  ABC-B'  is  equivalent  to  tri- 
angular prism  ADC-n'. 


268 


SOLID   GEOMETRY.— BOOK    VIIL 


Pass  a  plane  ±  to  AA^  so  as  to  form  a  rt.  section  ahcd  of 
the  parallelopiped,  and  intersecting  AG'  in  ac. 


Since  ac  is  a  diagonal  of  par'm  ahcd, 

A  ahc  =  A  adc. 

Now  ABC-B'  ^  a  rt.  prism  having  for  base  ~ 
A  abc,  and  alt.  =  BB' ; 

and   ADC-d'  ^  a  rt.   prism  having  for  base 
A  adc,  and  alt.  =  BB' ; 

.'.  prism  ABC-B'  =0=  prism  ADC-D',     q.e.d. 


(140) 


(519) 


(Ax.  1) 


(since  they  are  equivalent  to  equal  rt.  prisms.) 

524.  Cor.  Any  triangidar  prism  is  equivalent  to  one  half 
the  parallelopiped  having  the  same  altitude  and  a  base  twice 
as  great. 


Proposition^   VI.     Theorem. 

525.  Any  parallelopiped  is  equivalent  to  a  rect- 
angular parallelopiped  having  the  same  altitude 
and  an  equivalent  base. 


Given:  A  parallelopiped  ABCD-C\  having  an  altitude  H; 

To  Prove:  ABCD-c'  is  equivalent  to  a  rectangular  parallelo- 
piped having  a  base  equivalent  to  ABCDy  and  altitude  equal  to  H. 


PRISMS.  269 

Through  A  and  B  pass  planes  each  ±  to  AB. 

By  the  intersections  of  these  planes  with  the  faces,  or 
the  faces  produced,  a  new  parallelopiped,  ABcd-c',  will  be 
formed,  of  the  same  altitude  as  the  given  figure,  and  hav- 
ing an  equivalent  base  (Const.,  328). 

Now  ABcd-c'  ^ABCD-&  (519);  for  taking  AD'  as  the 
base  of  the  latter,  then  ABcd-c'  is  a  rt.  prism  whose  base, 
Ad*,  is  a  rt.  section  of  the  given  prism,  and  whose  altitude 
is  AB,  a  lateral  edge  of  the  same. 

Again  (Fig.  2),  through  the  edges  AB  and  dc  pass  planes 
Af,  dg,  1.  to  Ac,  the  base. 

By  the  intersection  of  these  planes  with  the  faces,  pro- 
duced if  necessary,  of  ABcd-c',  a  rectangular  parallelopiped, 
ABcd-g,  will  be  formed,  having  the  same  base  and  altitude 
as  ABcd-c',  or  ABh'a'-c'. 

Now,  by  regarding  Ah'  as  base  of  ABh'a'-c',  and  Af  as  a 
right  section,  it  is  seen  that 

ABcd-g  <>  ABcd-c'  (519).     Hence  (Ax.  1) 

ABCB-c'  ^  ABcd-g,  a  rect.  p'ped,  etc.,  q.e.d. 

(since  each  is  equivalent  to  ABcd-c'.) 

To  obviate  a  frequent  cause  of  difficulty  to  the  student, 
it  may  be  well  to  remark  that  the  necessity  of  the  double 
construction  originates  in  the  fact  that  the  given  solid 
ABCD-C'  being  any  parallelopiped,  we  have  to  assume  the 
possibility  that  both  pairs  of  faces,  AB'  and  DC',  AD'  and 
BC',  are  oblique  to  the  base  AC.  If,  in  the  given  solid,  we 
suppose  AB'  perpendicular  to  the  base,  then  ABcd-c',  the 
parallelopiped  first  constructed,  will  be  rectangular^  and  no 
further  construction  be  necessary. 

Exercise  730,  In  the  diagram  for  Prop.  VI.,  left-hand  figure, 
show  that  the  solid  AA'D'D-d'  is  equivalent  to  the  solid  BB'C^C-c'. 

731.  In  AA'D'D-d',  what  kinds  of  polyhedral  angles  are  those 
having  their  vertices  at  A  and  Z),  respectively  ? 


270 


SOLID   GEOMETRY.  — BOOK    VII I. 


Proposition  VII.     Theorem. 

526.  Rectangular  parallelopipeds  with  equal  bases 
are  to  each  other  as  their  altitudes. 


\L                 i^i\ 

\ 

\ 

N 

i  •■■.. 

E                           :■■••. 

X 

C'-... 

D\ 


F' 


\ 

L' 

=  \ 

"T--.. 

-\ 

i'--.. 

E               :•••.. 

\ 

C'-:. 

Given:  Two  rectangular  parallelopipeds,  AF,  a'f',  with  equal 
bases,  and  altitudes  AL,  a'l' ; 

To  Prove :     p'ped  AF  :  p'ped  a'f'  =  al  :  A'l'. 

1°.  When  AL  and  A'lJ  are  commensurable. 

Let  ^'J5;  be  a  common  measure  of  AL  and  A'l',  so  that 
A'f  can  be  laid  off  7  times  on  AL  and  5  times  on  A'L'. 
Through  the  points  of  division  pass  planes  J_  to  the  edges. 
Having  equal  bases  (510)  and  equal  altitudes,   (Const.) 
the  p'peds  thus  formed  are  equal.  (518) 

Since  AL:A'l'=7:5,  (Hyp.) 

and  p'ped  AF  :  p'ped  A'F' =7  :  5,  (Const.) 

p'ped^2^:p'ped^'j^'  =  ^i:^7.'.  q.e.d.  (P.  232'") 

2°.         When  AL  and  A^L'  are  incommensurable. 

Suppose  a'l'  divided  into  any  number  of  equal  parts,  n, 
and  that  AL  contains  m  such  parts  with  a  remainder  NL. 
Through  N  pass  a  plane  perpendicular  to  the  edges  and  cut- 
ting off  the  parallelopiped  AK, 


PRISMS. 


271 


Since  AN  and  A^L^  are  commensurable,     (Const.) 

AN  _  m  _  p'ped  AK  . 
A^L^~  n  "p'ped  ^'J^'' 

X 


(1°) 


and 


n 


n 


AF      m      ic' 

—  — I — 


A'F' 


n 


n 


since  AL  and  ^iT  are  slightly  greater  than  AN  and  AK  resp. 
Kow  when  n  is  taken  indefinitely  great,  -  and  -  become 


indefinitely  small ; 

p'ped  AF  _  AL 
p'ped  A'F'~  A'L'' 

(being  the  limits  of  variables  always  equal.) 


Q.E.D.     (255) 


527.  Cor.  If  two  rectangular  parallelopipeds  have  two 
dimensions  in  common,  they  are  to  each  other  as  their  third 
dimensions. 


Proposition  VIII.     Theorem. 

528.  Rectangular  parallelopipeds  having  equal  alti- 
tudes are  to  each  other  as  their  bases. 


Given :  Two  rectangular  parallelopipeds,  CA,  CD,  having  equal 
altitudes,  and  bases  B  G,  CE  ; 

To  Prove :  P'ped  CA  :  p'ped  CD  —  base  BG  :  base  CE. 


272 


SOLID   GEOMETRY.— BOOK    VIIL 


Place  the  parallelepipeds  so  that  the  edge  CF  may  be 

common,  and  the  right  dihedral  angles  at  CF  vertical. 

Produce  the  faces  AG,  BG,  AF,  Dl,  so  as  to  meet,  and 

form  a  third  rectangular  parallelopiped,  CH. 
Since  CA,  CH,  have  the  same  base,  FG,  and  the  altitudes 
BC,  CI, 

CA  :  CH  =  BC  :  CI  =  BC  '  CG  :  CI  -  CG.        (526,  318) 

Since  CD,  CH,  have  the  same  base,  FI,  and  altitudes  CK,  CG, 

CD  :  CH  =  CK  :  CG  =  CI  '  CK  :  CI  '  CG  ;     (526,  318) 

.-.  CA:  CD  =  rect.  BC  -  CG:  rect.  CI  •  CK ;  (249) 

i.e.,  p'ped  CA  :  p'ped  CD  =  base  BG  :  base  CE.        q.e.d. 

529.  Scholium.  The  foregoing  proposition  may  be  ex- 
pressed as  follows  : 

If  two  rectangular  parallelopipeds  have  one  dimension  in 
common,  they  are  to  each  other  as  the  products  of  their  other 
two  dimensions. 


Proposition  IX.     Theorem. 

530.  Rectangular  parallelopipeds  are  to  each  other 
as  the  products  of  their  three  dimensions. 


A 

^                       I 

^      A 

' i 

\        ^ 

i 

A      / 

i     p'     c 
/ 

r 

/ 

Given :  Two  rectangular  parallelopipeds,  P,  P',  with  dimensions 
a,  b,  c,  and  a',  b',  c',  respectively; 

To  Prove :        P  :  P'  =  a  x  b  x  c  :  a'  x  b'  x  c\ 


PRISMS.  •  273 

Let  Q  be  a  third  rectangular  parallelepiped  whose  dimen- 
sions are  a',  b,  c. 

Since  P,  Q,  have  two  dimensions  b,  c,  in  common,  (Hyp.) 

P:Q=a:a',   '  (527) 

Since  Q,  P',  have  the  dimension  a'  in  common,  (Hyp.) 

Q:P'  =  bxc:b'  xc'.  (529) 

.-.  P  :  P'  =  a  X  b  X  c  :  a'  X  b'  X  c'.  q.e.d.  (242) 

531.  Definition.  A  cube  is  a  rectangular  parallelopiped 
whose  faces  are  all  squares. 

CoR.    The  edges  of  a  cube  are  all  equal. 

532.  Definition.  The  U7iit  of  volume  is  the  cube  whose 
edge  is  the  linear  unit,  and  whose  base  is,  consequently,  the 
unit  of  area. 

533.  Cor.  1.  The  volume  of  a  rectangtdar  parallelopiped 
is  measured  by  the  product  of  its  three  dimeyisions. 

For  if  a,  b,  c,  be  the  dimensions  of  a  rectangular  paral- 
lelopiped P,  and  C7be  the  unit  cube,  then  (530), 

P:  U=abc:  1x1x1; 
.-.  P—Ux  abc. 

This  conld  be  expressed  at  greater  length  as  follows  : 
The  number  of  unit  cubes  in  any  rectangular  parallelopiped 
is  equal  to  the  number  of  units  in  the  product  of  the  numeri- 
cal measures  of  its  length,  breadth,  and  thickness. 

534.  Cor.  2.  The  volume  of  a  cube  is  measured  by  the 
cube  of  its  edge. 

Thus  if  the  edge  of  a  cube  be  7  linear  units,  the  cube 
contains  7^  =  343  unit  cubes ;  if  the  edge  be  a  linear  units, 
the  cube  contains  a^  times  the  unit  cube. 
Geom.  — 18 


274  SOLID   GEOMETRY.  — BOOK  VIII. 

Proposition  X.     Theorem. 

535.  The  volume  of  any  prism  is  measured  hy  the 
product  of  its  base  and  altitude. 

1°.  Any  parallelopiped  is  equivalent  to  a  rectangular 
parallelopiped  having  the  same  altitude  and  an  equivalent 
base  (525)  ;  and  the  volume  of  the  latter  is  measured  by 
the  product  of  its  three  dimensions,  —  that  is,  of  its  base 
and  altitude  (533)  ;  hence  the  volume  of  any  parallelopiped 
is  measured  by  the  product  of  its  base  and  altitude. 

2°.  Any  triangular  prism  is  equivalent  to  one  half  the 
parallelopiped  having  the  same  altitude  and  a  base  of  twice 
the  area  (524)  ;  now,  the  volume  of  the  latter  being  meas- 
ured by  the  product  of  its  base  and  altitude  (1°),  the  volume 
of  the  triangular  prism  is  also  measured  by  the  product  of  its 
base  and  altitude. 

S°.  By  passing  planes  through  its  lat- 
eral edges,  any  prism  can  be  divided  into 
triangular  prisms  whose  altitudes  are  the 
same  as  that  of  the  given  prism,  and  whose 
triangular  bases  together  form  the  base 
of  the  given  prism.  As  the  volume  of 
each  of  these  triangular  prisms  is  meas- 
ured by  the  product  of  its  base  and  altitude  (2°),  the  volume 
of  any  prism  is  measured  by  the  product  of  its  base  a7id 
altitude,     q.e.d. 

536.  Cor.  Prisms  having  equivalent  bases  are  to  each 
other  as  their  altitudes;  prisms  having  equal  altitudes  are  to 
each  other  as  their  bases ;  and  prisms  are  to  each  other  as  the 
products  of  their  bases  and  altitudes. 

Exercise  732.  Two  triangular  prisms,  A  and  _B,  liave  tlie  same  alti- 
tude. A  has  for  base  a  right-isosceles  triangle  ;  B,  for  base  an  equi- 
lateral triangle  of  side  equal  to  the  hypotenuse  of  the  base  of  A.  Find 
the  ratio  of  the  volume  of  A  to  that  of  B.- 

733.  Find  the  ratio  of  the  lateral  area  of  A  to  that  of  B. 


PYRAMIDS,  275 

PYRAMIDS. 

537.  A  pyramid  is  a  polyhedron  bounded  by  a  polygon 
called  the  base,  and  by  triangular  planes  meeting  in  a  com- 
mon point  called  the  vertex.  A  plane  intersecting  the  faces 
of  any  polyhedral  angle  cuts  off  a  pyramid. 

The  terms  lateral  face,  lateral  surface,  lateral  edge,  basal 
edge,  are  defined  as  for  prisms  (504). 

538.  The  altitude  of  a  pyramid  is 
the  perpendicular  distance  from  its 
vertex  to  the  base ;  as  SP. 

539.  A  regular  pyraynid  has  for 
base  a  regular  polygon,  and  has  its 
vertex  in  the  perpendicular  at  the 
center  of  the  base,  which  perpendicu- 
lar is  called  the  axis  of  the  pyramid. 

540.  The  slant  height  of  a  regular  pyramid  is  the  alti- 
tude of  any  lateral  face. 

541.  A  pyramid  is  triangular,  quadrangular,  pentagonal, 
etc.,  according  as  its  base  is  a  triangle,  quadrangle,  p)enta- 
gon,  etc.  In  the  triangular  pyramid,  or  tetrahedron  (499), 
any  one  of  the  faces  may  be  regarded  as  the  base. 

542.  A  truncated  pyramid  is  the  portion  /C^^nA 
of  a  pyramid  included  between  the  base  and        /  /     JA  \ 
a  plane  that  intersects  all  the  lateral  faces.       \  /         \  y 

543.  A  frustum  of  a  pyramid  is  a  trun- 
cated pyramid  in  which  the  intersecting  plane  is  parallel  to 
the  base.     The  base  of  the  pyramid  is  called  the  lower  base 
of  the  frustum ;  the  parallel  section,  the  upper  base. 

544.  The  altitude  of  a  frustum  is  the  perpendicular  dis- 
tance between  its  bases ;  the  slant  height  of  a  frustum  of  a 
regular  pyramid  is  the  altitude  of  any  lateral  face. 


276  SOLID   GEOMETRY.- BOOK   VIIL 

Proposition  XI.     Theorem. 

545.  If  a  pyramid  be  cut  by  a  -plane  parallel  to 
the  base : 

1°.  The  edges  and  altitude  will  be  divided  pro- 
portionally. 

2°.  The  section  is  a  polygon  similar  to  the  base. 


Given :  A  pyramid  S-ABD,  whose  altitude  SP  is  cut  in  p  by 
a  plane  abd  parallel  to  the  base ; 

To  Prove :  1°,  SA  :  Sa  =  SB  :  Sb  =  SP  :  SjJ,  etc. ; 
2°,  abd  is  similar  to  ABD. 

1°.    Suppose  a  plane  passed  through  S  II  to  ABD. 
Since  the  edges  and  altitude  are  cut  by  II  planes, 

(Hyp.  and  Const.) 

SA  :  Sa  =  SB  :  Sb  =  SC  :Sc  =  SP:  Sp,  etc.     q.e.d.     (459) 

2°.  Since  plane  abd  is  II  to  plane  ABD,  (Hyp.) 

ab  is  II  to  AB,  be  is  II  to  BC,  cd  is  II  to  CD,  etc.,     (451) 

and  they  are  similarly  directed,  (H^) 

abd  and  ABD  are  mutually  equiangular.        (455) 

Since  ab  is  II  to  AB,  and  be  is  II  to  BC, 

A  Sab  is  sira.  to  A  SAB,  and  A  sbc  to  A  SBC;       (291) 

.-.  ab:AB  =  Sb:  SB,  and  be:  BC  =  Sb:  SB  ;       (284) 

.-.  ab:AB  =  bc:BC. 


PYRAMIDS.  277 

In  the  same  way  we  show  that 
be :  BC  =  cd:  CD  =  de:  BE,  etc. ; 
.-.  abode  is  similar  to  ABCBE.     q.e.d.     (284) 

546.  Cor.  1.  The  area  of  any  section  of  a  pyramid  paral- 
lel to  the  base,  is  proportional  to  the  square  of  its  distance  from 
the  vertex. 

For  parallel  sections  being  similar  to  the  base  (545),  their 
areas  are  proportional  to  the  squares  of  their  homologous 
sides  (344).     Thus 

abd :  ABB  =  «6^ :  AB^  =  Ja  :SA^  =^Sp  :  JP. 

547.  Cor.  2.  If  two  pyramids,  S-ABB,  s'-A^b'b',  having 
equal  altitudes,  SF,  s'p',  are  cut  by  planes  parallel  to  their 
bases,  and  at  equal  distances,  Sp,  s'p',  from  their  vertices,  the 
sections,  abd,  a'b'd',  will  be  to  each  other  as  the  bases. 


For  abd  :  ABB  =  Sp  :  SP",       ) 
and  a'b'd' :  A'b'b'=  s'p'  :  s'p'  ;  \ 


(546) 


but  Sp  =  S'p'  and  SP  =  s'p' ;  (Hyp. ) 

.-.  abd  :  ABB  =  a'b'd' :  A'b'b'. 

548.  CoR.  3.  If  tivo  pyramids  have  equal  altitudes  and 
equivalent  bases,  sections  made  by  planes  parallel  to  their 
bases,  and  at  equal  distances  from  the  vertices,  are  equivalent. 

Exercise  734.  Show  that  a  plane  perpendicular  to  the  axis  of  a 
regular  pyramid  forms  equal  dihedral  angles  with  all  the  faces  of  the 
pyramid. 

735.  In  order  that  a  plane  intersecting  the  faces  of  a  polyhedral 
angle  may  cut  off  a  regular  pyramid,  what  conditions  must  be  fulfilled 
in  regard  to  the  form  of  the  polyhedral  and  the  inclination  of  the  plane  ? 

736.  In  the  diagram  for  Prop.  XI.,  if  a  plane  be  passed  through  the 
mid  point  t)f  pP,  parallel  to  the  base,  show  that  the  perimeter  of  the 
section  thus  formed  will  be  equal  to  half  the  sum  of  the  perimeters  of 
ABD  and  abd. 


278  SOLID   GEOMETRY.  — BOOK   VIIL 

Proposition  XII.     Theorem. 

549.  The  lateral  surface  of  a  regular  pyramid  is 
equivalent  to  one  half  the  j^ectangle  contained  hy  its 
perimeter  and  slant  height. 


A  B 

Given :  A  regular  pyramid  S-ABD,  and  SF  its  slant  height ; 
To  Prove :  The  lateral  surface  of  S-ABD  is  equivalent  to 
J-rect.  Si^-  {AB-{-BC-\ EA). 

Since  ABD  is  a  regular  polygon,  (Hyp.) 

AB  =BC^CD  =DE  =  EA;  (370) 

since  A,  B,  C,  D,  E,  are  equally  distant  from  P,  (539) 

SA  =  SB  =  SC  =  SD  =  SE;  (437) 

.-.  isos.  A5fyli?=isos.A,Si?(7=isos.  A-SCZ),  etc.  (69) 

But  A  SAE  o=  I  rect.  SF  •  AE ;     .  (331) 

.-.  ASAB  ^  ASBC  -\ A  SAE 

^^SF  -(AB  -^BC+-"AE);  (335) 

i.e.,  lat.  surf,  of  S-ABD  ^  }j  SF  '  perimeter.  q.e.d. 

550.  Cor.  1.  The  lateral  surface  of  a  frustum,  of  a 
regular  pyramid  is  equivaleyit  to  one  half  the  rectangle  con- 
tained hy  the  slant  height  of  the  frustum  and  the  sum  of  the 
perimeters  of  the  bases.  (338) 

For  it  is  the  sum  of  as  many  trapezoids  as  the  base  has 
sides,  having  for  common  altitude  the  slant  height  of  the 
frustum.  (544) 

551.  Cor.  2.  The  dihedral  and  trihedral  angles  at  the  base 
of  a  regular  pyramid  are  all  equal. 


PYRAMIDS. 


279 


Proposition  XIII.    Theorem. 

552.  Triangular  pi/ramids  having  equivalent  bases 
and  equal  altitudes  are  equivalent. 


1 

k 

g'L- 

:rj^ 

■■■■■J^ 

4.\^l 

^^     /  \ 

^^^^^^^ 

Given:  Two  triangular  pyramids,  S-ABC  and  s'-a'b'c',  with 
equivalent  bases,  ABC  and  a'b'c',  and  the  same  altitude  AL; 
To  Prove :  S-ABC  is  equivalent  to  s'-a'b'c'. 

Place  the  pyramids  so  that  they  shall  be  in  the  same 
plane,  and  AL  be  their  common  altitude. 

Divide  AL  into  n  equal  parts,  Aa,  etc.,  and  through  the 
points  of  division  pass  planes  parallel  to  the  plane  of 
the  bases. 
The  corresponding  sections  thus  formed  are  equivalent ;  (548) 
that  is,  DEF  ^  d'e'f',  ghi  ^  g'h'T,  etc. 

On  the  triangles  ABC,  DEF,  etc.,  as  lower  bases,  construct 
prisms  D-ABC,  G-DEF,  etc.,  whose  lateral  edges  are 
parallel  to  SA,  and  whose  altitudes  are  each  equal 
to  Aa. 

On  the  triangles  d'e'f',  g'h'i',  etc.,  as  upper  bases,  con- 
struct prisms  a'-d'e'f',  d'-g'h'i',  etc.,  whose  lateral 
edges  are  parallel  to  S'a',  and  whose  altitudes  are  each 
equal  to  Aa. 


280  SOLID   GEOMETRY.  — BOOK    VIII. 

Now  prism  G^-D^i^  =c=  prism  A'-d'e'f^  (536) 

because  they  have  equivalent  bases  and  the  same  altitude. 
Similarly,  K-GHI  o  d'-g'h'i',  etc. 

That  is,  corresponding  to  each  triangular  prism  con- 
structed upon  a  section  of  S-ABC  as  lower  base,  is  an  equiva- 
lent prism  constructed  upon  the  corresponding  section  of 
S'-A'b'c'  as  upper  base.  Hence,  the  sum  of  all  the  prisms 
circumscribing  S-ABC  differs  from  the  sum  of  the  prisms 
inscribed  in  s'-a'b'c'  by  the  prism  D-ABC. 

But  the  sum  of  all  the  prisms  circumscribing  S-ABC  is 
greater  than  that  pyramid ;  and  the  sum  of  all  the  prisms 
inscribed  in  S'-A'b'c'  is  less  than  S'-A'b'c'.  Hence,  these 
pyramids  differ  in  volume  by  a  volume  less  than  prism 
D-ABC. 

Now  if  n  be  taken  indefinitely  great,  the  altitude  Aa 
becomes  infinitesimal,  and  therefore  the  prism  D-ABC 
becomes  infinitesimal.  Hence,  as  they  cannot  differ  by 
even  an  infinitesimal  volume, 

pyramid  S-ABC  ^  pyramid  S'-A'b'c'.  q.e.d. 


Exercise  737.  In  the  diagram  for  Prop.  XIL ,  if  the  axis  is  equal 
to  the  apothem  of  the  base,  what  is  the  inclination  of  each  face  to  the 
base  ? 

738.  In  the  same  diagram,  if  the  base  is  a  regular  pentagon,  and 
Z  SAB  =  70°,  what  is  the  sum  of  the  face  angles  ? 

739.  In  the  same  diagram,  show  that  the  mid  point  of  SF  is  equi- 
distant from  S  and  P. 

740.  In  a  regular  pyramid,  the  sum  of  the  squares  of  the  lateral 
edges  is  equivalent  to  one  fourth  of  the  sum  of  the  squares  of  the 
basal  edges,  and  n  times  the  square  of  the  slant  height. 

741.  The  perpendicular  from  the  foot  of  the  axis  of  a  regular  pyra- 
mid to  the  slant  height  is  a  mean  proportional  between  the  segments 
into  which  it  divides  it. 

742.  The  axis  and  the  slant  height  of  a  regular  pyramid  being 
given,  find  the  apothem  of  the  base. 


PYRAMIDS.  281 

Proposition  XIV.     Theorem. 

553.  A  triangular  pyramid  is  one  third  of  a  tri- 
angular prism  having  the  same  base  and  altitude. 


Given:  A  triangular  pyramid  B'-ABC,  and  a  triangular  prism 
ABC-C',  on  the  same  base. 

To  Prove :  B'-ABC  is  equivalent  to  ^  ABC-C'. 

Take  away  the  pyramid  B'-ABC;  there  remains  the 
quadrangular  pyramid  whose  vertex  is  B'  and  whose  base 
is  the  parallelogram  AC'. 

Through  B',  A',  C,  pass  a  plane.  It  will  divide  the  pyra- 
mid b'-AA'c'C  into  two  triangular  pyramids,  which  are 
equivalent  (552)  since  the  bases  are  halves  of  the  parallelo- 
gram A  c',  and  they  have  the  same  altitude,  —  the  perpen- 
dicular from  b'  upon  the  base  AC'. 

But  pyramid  b'-A'c'C,  or  C-A'b'c',  ^  pyramid  B'-ABC 
(552);  for  they  have  equal  bases,  A'b'c',  ABC  (503),  and 
the  same  altitude,  namely,  that  of  the  prism.     Hence 

ipjT.  B'-ABC  +  pyr.  B'-AA'C 

4-  pyr.  B'-A'c'C  o  3  pyr.  B'-ABC ; 
i.e.,  prism  ABC-C'  o=  3  pyr.  B'-ABC.  q.e.d. 

554.  OoR.  The  volume  of  a  triavgular  pyramid  is  meas- 
ured by  one  third  the  product  of  its  base  and  altitude. 


282  SOLID   GEOMETRY.  — BOOK    VIII. 

Proposition  XV.     Theorem. 

555.  Any  pyramid  is  one  third  of  a  prism  having 
the  same  hase  and  altitude. 


Given :  Any  pyramid  S-ABD,  with  base  JBD  and  altitude  SP  ; 
To  Prove:  S-ABD  is  equivalent  to  one  third  the  prism  on 
base  ABD  with  altitude  SP. 

Througli  SA  pass  planes  SAD,  SAC,  etc.  These  planes 
divide  S-ABD  into  triangular  pyramids,  whose  bases  make 
up  the  base  ABD,  and  whose  common  altitude  is  SP. 

Each  of  these  triangular  pyramids  is  one  third  of  the 
triangular  prism  with  altitude  SP  that  could  be  constructed 
on  the  same  base  (553). 

Hence  the  sum  of  the  triangular  pyramids  that  make  up 
the  given  pyramid,  is  one  third  of  the  prism  with  altitude 
SP  that  could  be  constructed  on  the  ba&e  ABD.  q.e.d. 

Scholium  1.  The  proposition  could  be  expressed  under 
the  form :  The  volume  of  any  pyramid  is  measured  by  one 
third  the  product  of  its  base  and  altitude. 

556.  Scholium  2.  The  volume  of  any  polyhedron  may  be 
found  by  dividing  it  into  prisms  or  pyramids,  and  comput- 
ing the  sum  of  their  volumes. 

557.  Cor.  Pyramids  of  equivalent  bases  are  as  their  alti- 
tudes; pyramids  of  equal  altitudes  are  as  their  bases;  any 
two  pyramids  are  as  the  products  of  their  bases  and  altitudes. 


PYRAMIDS.  283 

Proposition  XVI.     Theorem. 

558.  A  frustum  of  a  triangular  pyramid  is  equiv- 
alent to  the  sum  of  three  pyramids  of  the  same  alti- 
tude as  the  frustuTVii  and  whose  bases  are  those  of  the 
frustum  and  a  mean  proportional  between  them. 


G 

c 

Given:  A  frustum  D-ABC,  with  bases  ABC,  DEF; 

To  Prove:  D-ABC  is  equivalent  to  three  pyramids  having 
the  altitude  of  the  frustum,  and  having  as  bases  ABC,  DEF, 
and  a  mean  proportional  between  ABC  and  DEF. 

Pass  planes  through  the  points  F,  A,  B,  and  through  i^,  D,  B. 
We  thus  divide  the  frustum  into  three  triangular  pyra- 
mids, ^ 

F-ABC,  B-DEF,  and  F-ADB. 

The  first  two  of  these  evidently  have  the  same  altitude  as 
the  frustum,  and  have  for  bases  ABC  and  DEF  respectively. 
The  pyramid  F-ADB,  we  shall  show,  is  equivalent  to  a  trian- 
gular pyramid  having  the  same  altitude  as  the  frustum,  and 
a  base  that  is  a  mean  proportional  between  ABC  and  DEF. 

In  the.  plane  DC,  draw  i^G^  ||  to  DA,  and  pass  a  pjane 
through  FOB. 

FGi^W  to  plane  ABED,  (448) 

whence  F  and  G  are  equally  distant  from  that  plane ; 

.  •.   py r.  G-A  DB=o  F-A  D  B.  (552) 

If,  again,  we  take  D  as  the  vertex,  and  A  GB  2is  the  base  of 
G-ADB,  its  altitude  is  the  same  as  that  of  the  frustum. 


284  SOLID   GEOMETRY.  — BOOK   VIII. 

Draw  GH  II  to  CB.     Then  Z  A  GH=  Z  dfe, 

and  Zgah=Zfde.  (455) 

Since  ^ i^  is  a  par'm,  AG  =  DF; 
.:  AAHG=ADEF  (63),  Siiid  AH=DE. 
Since  A  ABC  and  ABG,  Aabg  and  AHG,  have  equal  alts., 
AABC:AABG=:AC:AG  =  AC:DF,} 
2indAABG:AAHG  =  AB:AH=AB:DE.\  ^        ^ 

But  AC:DF=zAB:DE, 
{A  ABC  and  DEF  being  similar.)  (545") 

.-.   ABC:  ABG  =  ABG:  AHG  =  ABG:  DEF;     (232'") 
I.e.,  A  ^5  G  is  a  mean  prop,  between  A  ABC  and  DEF. 

569.  Cor.  1.  T/ie  volume  of  a  frustum  of  a  triangular 
pyramid  is  measured  by  the  product  of  one  third  its  altitude 
into  the  sum  of  its  bases  and  a  mean  proportional  between 
them. 

560.  Cor.  2.  The  volume  of  a  frustum  of  any  pyramid  is 
measured  by  the  product  of  one  third  its  altitude  into  the  sum 
of  its  bases  and  a  mean  proportional  between  them. 

For  planes  that  divide  the  complete  pyramid  into  trian- 
gular pyramids  will  divide  the  frustum  into  triangular  frus- 
tums having  the  altitude  of  the  given  frustum.  If,  now,  a 
plane  be  passed  so  as  to  cut  the  given  frustums  in  a  section 
that  is  a  mean  proportional  between  the  bases,  it  will  cut 
each  triangular  frustum  in  a  section  that  is  a  mean  propor- 
tional between  its  upper  and  lower  bases ;  that  is,  the  mean 
proportional  between  the  bases  of  the  given  frustum  is  the 
sum  of  those  between  the  bases  of  the  triangular  frustums. 
The  volume  of  the  given  frustum  is  the  sum  of  the  volumes 
of  the  triangular  frustum ;  hence  it  is  measured  by  the 
product  of  one  third  their  common  altitude  into  the  sum  of 
the  upper  and  lower  bases  of  the  •  triangular  frustums  and 
the  mean  proportionals  between  them. 


PYRAMIDS. 


285 


Proposition  XVII.     Theorem. 

561.  Tetrahedrons  with  a  trihedral  angle  of  the  one 
equal  to  a  trihedral  angle  of  the  other,  are  to  each 
other  as  the  products  of  the  edges  of  these  trihedral 
angles. 


Given :   V  and  F',  the  volumes  of  two  tetrahedrons  having  trihe- 
dral angle  A  of  the  one  equal  to  trihedral  angle  A'  of  the  other ; 
To  Prove :    V:  V'  =  AB  X  AC  X  AS:  A'b'  X  A'c'  X  A's'. 

Apply  one  tetrahedron  to  the  other  so  that  A'  =^  A. 
From  S  and  -S'  draw  SP  and  S'p',  Js  to  ABC  Sind  A'b'c', 
"   and  let  their  plane  intersect  ABC  in  AP'f. 
V:  V"  =  ABC  X  SP:  A'b'C'  X  S'P'. 


But  ABC:  A'B'C'  =  AB  X  AC:  A'b'  X  A'C', 


(557) 
(341) 
(289) 
(288) 


^i\d  SP:  S'P'  =  AS:  A's', 
(since  A  SAP  is  similar  to  A  S'A'P'.) 
Therefore,  making  the  proper  substitutions,  we  have, 

V:  V'  =  AB  X  AC  X  AS:  A'b'  X  A'C'  X  A'S'.        q.e.d. 


562.   Cor.  Similar  tetrahedrons  are  as  the  cubes  of  their 
homologous  edges. 

For  let  S-ABC,  s'- a'b'c'  be  the  similar  tetrahedrons. 

Since  V:  V'=  SAXSBX  SC:S'A'  X  S'b'  X  S'c',      (561) 
and  SA  :  S'A'  =  SB  :  S'b'  =  SC:S'C',  (Hyp.) 

V:r'  =  SAX  SAXSA:  S'A'  X  S'A'  X  S'A'  =  'SA^  :  S^l 


286  SOLID   GEOMETRY.  — BOOK    VIII. 

THE  REGULAR  POLYHEDRONS. 

563.  Definition.  A  regular  polyhedron  has  its  faces  all 
equal,  regular  polygons,  and  its  polyhedral  angles  all  equal. 

Proposition  XVIII.     Theorem. 

564.  There  are  only  five  possible  regular  polyhedrons. 

The  faces  of  a  regular  polyhedron  must  be  regular  poly- 
gons (563)  ;  at  least  three  are  necessary  to  form  each  poly- 
hedral angle  (487)  ;  and  the  sum  of  the  face  angles  of  each 
polyhedral  angle  must  be  less  than  four  right  angles  (492). 

1°.  The  simplest  regular  polygon  is  the  equilateral  trian- 
gle, each  of  whose  angles  is  60°.     Now 

60°  X  3  =  180°,  60°  X  4  =  240°,  and  60°  x  5  =  300° ; 
but  60°  X  6  =  360°  =  4  rt.  A.      Hence   only   three   regular 
polyhedrons  can  have  equilateral  triangles  as  faces. 

2°.  The  regular  four-sided  polygon,  or  square,  has  each  of 
its  angles  90°.     Now 

90°  X  3  =  270°,  but  90°  X  4  =  360°  =  4  rt.  A. 
Hence  only  one  regular  polyhedron  can  be  formed  having 
squares  as  faces. 

3°.  The  regular  pentagon  has  each  angle  108°.     Now 
108°  X  3  =  324°,  but  108°  x  4  =  432°  >  4  rt.  A. 
Hence  only  one  regular  polyhedron  can  be  formed  having 
pentagons  as  faces. 

4°.  The  regular  hexagon  has  each  of  its  angles  120°. 
Now  120°  X  3  =  360°  =  4  rt.  A.  Hence  no  regular  polyhe- 
dron can  be  formed  having  as  faces  regular  polygons  of 
six  or  more  sides.  Hence  only  five  regular  polyhedrons  can 
be  formed:  three  having  equilateral  triangles  as  faces, — 
namely,  the  tetrahedron,  the  octahedron,  and  the  icosahedron ; 
one  having  squares  as  faces,  —  the  hexahedron  or  cube;  and 
one  having  pentagons  as  faces,  —  the  dodecahedron. 


THE  REGULAR   POLYHEDRONS. 


287 


Proposition  XIX.     Problem. 

565.  To  construct  the  regular  polyhedrons,  an  edge 
being  given. 

Given :  A  straight  line  AB  slb  edge ; 

Required :   To  construct  the  regular  polyhedrons. 


1°.  Upon  AB  construct  an  equilateral  triangle  ABC,  and 
find  its  center  0.  At  0  draw  OD  ±  to  AB  C,  and  in  OD  take 
a  point  D  such  that  AB  =  AB.  Join  BA,  BB,  BC ;  B-ABC  is 
the  required  tetrahedron. 

For  the  four  faces  are  equilateral  triangles  (Const.)  ;  and 
the  trihedral  angles  A,  B,  C,  B,  are  equal  (495),  since  their 
faces  are  all  equal. 


H c 

B 


2°.  Upon  AB  construct  a  square  AC,  and  upon  the  sides 
of  this  square  construct  squares  AF,  BG,  CH,  BE,  in  planes 
perpendicular  to  the  plane  of  AC ;  AG  is  the  required  cube. 


288 


SOLID   GEOMETRY.  — ROOK    VIIL 


For  the  six  faces  are  squares  (Const.),  and  the  trihedral 
angles  A,  B,  C,  D,  E,  F,  G,  H,  are  equal  (495),  since  their 
faces  are  all  equal. 


3°.  Upon  ^J5  construct  a  square  AC,  and  at  its  center  0, 
draw  EF  JL  to  plane  AC ;  make  OF  =  OE  =  OA,  and  join 
EA,  EB,  EC,  ED,  FA,  FB,  FC,  FD.  These  edges  are  equal  to 
each  other  and  to  OA,  since  AGE,  AGF,  etc.,  are  equal  right 
triangles;  hence  the  faces  of  the  figure  are  equal  equilateral 
triangles.  Also,  since  the  triangles  DEB,  DFB,  are  equal, 
DEBF  is  a  square;  whence  it  follows  that  the  pyramid 
A-DEBF  is  equal  to  the  pyramid  E-ABCD  ;  hence  the  poly- 
hedral angles  A  and  E  are  equal ;  hence  all  the  polyhedral 
angles  of  the  figure  are  equal,  and  the  figure  is  a  regular 
octahedron. 


4°.  Upon  AB  construct  a  regular  pentagon  ABODE,  and 
to  each  side  of  ABODE  apply  an  equal  pentagon,  so  inclined 
to  the  plane  of  ABCDE  as  to  form  trihedral  angles  at  A,  B, 


THE  REGULAR  POLYHEDRONS. 


289 


C,  D,  E.  We  thus  obtain  a  convex  surface  FHLNP,  com- 
posed of  six  regular  pentagons.  Construct  a  second  con- 
vex surface  f'h'l'n'p',  equal  to  the  first,  and  apply  it  so  as 
to  form  a  single  convex  surface.  This  will  contain  the 
figure  required. 

For  the  faces  are  all  equal  pentagons  (Const.),  and  the 
trihedral  angles  are  equal,  being  contained  by  equal  faces, 
and  there  being  twelve  such  faces,  the  figure  is  a  dodeca- 
hedron. 


5°.  Upon  AB  construct  a  regular  pentagon  ABODE,  and 
at  its  center  0,  draw  OS  ±  to  abode.  Take  s  so  that 
SA==AB,  and  join  SA,  SB,  etc.,  thus  obtaining  a  regular 
pyramid  S-ABCDE,  having  each  of  its  faces  an  equilateral 
triangle.  Now  take  the  vertices  A  and  B  as  the  vertices 
of  two  other  pyramids  A-BSEFG  and  B-ASOHG,  having  in 
common  with  S-ABODE  the  faces  ASB,ASE,  and  ^<S5,  BSO 
respectively,  and  in  common  with  each  other,  the  faces  ASB 
and  ABQ.  We  thus  obtain  a  convex  surface  CDEFGH, 
consisting  of  ten  equal  equilateral  faces. 

Construct  a  second  convex  surface  O'd'e'f'g'h',  equal  to 
the  first,  and  apply  it  to  the  first  so  as  to  form  a  single  con- 
vex surface.     This  will  contain  the  figure  required. 

For  any  two  consecutive  face  angles  of  the  one  will  unite 
with  any  three  consecutive  face  angles  of  the  other  so  as  to 
form  a  regular  pentahedral  angle.  Hence  the  pentahedral 
angles  are  all  equal,  being  formed  by  equal  faces,  and  there 
are  twenty  such  faces ;  hence  they  form  an  icosahedron. 
Geom.  — 19 


290 


SOLID   GEOMETRY.  — BOOK    VIII. 


Scholium.  The  regular  polyhedrons  may  be  constructed 
in  the  following  way : 

Having  drawn  on  cardboard  the  diagrams  below,  cut 
through  the  heavy  lines  and  half  through-  the  dotted  lines ; 


Tetrahedron 


Octahedron 


Hexahedron 


Dodecahedron 


Icosahedron 


bring  the  edges  together,  and  keep  them  in  position  by 
pasting  over  them  strips  of  paper. 


EXERCISES, 


QUESTIONS. 


743.  What  is  the  least  number  of  points  that  can  limit  a  line  ?  The 
least  number  of  straight  lines  that  can  limit  a  surface  ?  The  least 
number  of  plane  surfaces  that  can  limit  a  solid  ? 

744.  What  is  the  lateral  surface  of  a  prism  whose  lateral  edge  is 
25  in.,  and  the  perimeter  of  whose  right  section  is  42  in.  ? 

745.  What  is  the  volume  of  a  right  prism  whose  altitude  is  40  in. , 
and  whose  base  contains  ^\  sq.  ft.  ? 

746.  With  what  theorem  of  Book  V.  is  Prop.  IV.  analogous  ? 

747.  With  what  theorems  of  Book  I.  are  Prop.  V.  and  its  corollary- 
analogous  ? 


EXERCISES.  291 

748.  With  what  theorems  of  Book  V.  are  Props.  VI.,  VII.,  VIII., 
and  IX.  analogous  ? 

749.  How  many  square  feet  are  there  in  the  surface  of  a  rectangular 
parallelopiped  whose  dimensions  are  16  in.,  22  in.,  and  30  in.,  respec- 
tively ? 

750.  How  many  cubic  feet  in  the  volume  of  the  same  solid  ? 

751.  The  dimensions  of  one  rectangular  parallelopiped  are  2  ft., 
5  ft.,  and  14  ft.,  respectively  ;  those  of  another  are  3  ft. ,  4  ft.,  and  10 
ft.,  respectively.     What  is  the  ratio  of  the  first  solid  to  the  second  ? 

752.  The  volume  of  a  rectangular  parallelopiped  is  96  cu.  ft.,  and  its 
altitude  is  8  ft.  3  in.     What  is  the  area  of  its  base  ? 

753.  The  volume  of  a  rectangular  parallelopiped  is  120  cu.  ft.,  and 
the  dimensions  of  its  base  are  4  ft.  5  in.  and  6  ft.  respectively.  What 
is  its  altitude  ? 

754.  The  edge  of  a  cube  is  4|  in.  What  are  its  volume  and  its 
entire  surface  ? 

755.  What  should  be  the  edge  of  a  cubical  box  that  shall  contain  a 
gallon  dry  measure  ? 

756.  What  should  be  the  edge  of  a  cube  so  that  its  entire  surface 
shall  be  a  square  foot  ? 

757.  What  should  be  the  altitude  of  a  prism  whose  base  is  3|  sq.  in. , 
so  that  it  may  have  the  same  volume  as  a  prism  whose  altitude  is 
6 1  in.  and  whose  base  is  2|  sq.  in.  ? 

758.  The  base  of  a  pyramid  is  12  sq.  ft.  and  its  altitude  is  6  ft. 
What  is  the  area  of  a  section  parallel  to  the  base  and  2  ft.  from  it  ? 

759.  The  perimeter  of  the  base  of  a  pyramid  is  15  in.;  its  slant 
height  is  7  in.     What  is  the  lateral  surface  ? 

760.  What  is  the  volume  of  a  pyramid  whose  base  is  12.3  sq.  in. ,  and 
whose  altitude  is  5.72  ft.? 

761.  The  bases  of  two  pyramids  are  6.4  sq.  ft.  and  8.1  sq.  ft.  re- 
spectively;  their  altitudes  are  9  in.  and  8  in.  respectively.  What 
is  their  ratio  ? 

762.  The  bases  of  a  frustum  of  a  pyramid  are  8  sq.  ft.  and  4| 
sq.  ft.  respectively,  and  its  altitude  is  5  ft.     What  is  its  volume  ? 

763.  The  bases  of  a  frustum  of  a  pyramid  are  20  sq.  in.  and  7.2 
sq.  in.  respectively.    Its  volume  is  400  cu.  in.    What  is  its  altitude  ? 


292  SOLID   GEOMETRY.  — BOOK    VIII. 


THEOREMS. 

764.  Every  section  of  a  prism  by  a  plane  parallel  to  the  lateral  edges 
is  a  parallelogram. 

765.  The  lateral  areas  of  right  prisms  having  equal  altitudes  are  as 
the  perimeters  of  their  bases. 

766.  The  diagonals  of  a  rectangular  parallelopiped  are  equal. 

767.  The  square  of  a  diagonal  of  a  rectangular  parallelopiped  is 
equal  to  the  sum  of  the  squares  of  the  three  diagonals  meeting  in  any 
vertex. 

768.  The  volume  of  a  triangular  prism  is  equal  to  one  half  the  prod- 
uct of  any  lateral  face  by  its  distance  from  the  opposite  edge. 

769.  The  volume  of  any  prism  is  equal  to  the  product  of  its  right 
section  by  an  edge. 

770.  The  four  diagonals  of  a  parallelopiped  bisect  each  other. 

771.  If  the  four  diagonals  of  a  four-sided  prism  pass  through  a 
common  point,  the  prism  is  a  parallelopiped. 

772.  Any  straight  line  passing  through  the  center  of  a  parallelopi- 
ped and  terminated  by  two  faces,  is  bisected  at  the  center. 

N.B.  —  The  center  of  a  parallelopiped  is  the  point  of  intersection  of  its 
diagonals. 

773.  Any  plane  passing  through  the  center  of  a  parallelopiped 
divides  it  into  two  equal  solids. 

774.  If  any  two  nonparallel  diagonal  planes  of  a  prism  are  perpen- 
dicular to  the  base,  the  prism  is  a  right  prism. 

775.  The  lateral  surface  of  any  pyramid  is  greater  than  its  base. 

776.  The  mid  points  of  the  edges  of  a  regular  tetrahedron  are  at  the 
vertices  of  a  regular  octahedron. 

777.  The  section  of  a  triangular  pyramid  made  by  a  plane  passed 
parallel  to  two  opposite  edges  is  a  parallelogram. 

778.  The  section  of  a  regular  tetrahedron  made  by  a  plane  passed 
parallel  to  two  opposite  edges  is  a  rectangle. 

779.  The  altitude  of  a  regular  tetrahedron  is  equal  to  the  sum  of 
the  perpendiculars  to  the  faces  from  any  point  within  the  figure. 


Book  IX. 

THE   THREE  ROUND  BODIES. 


3»iC 


Of  the  solids  that  are  bounded  by  curved  surfaces,  only 
three  are  treated  of  in  Elementary  Geometry,  viz.,  the  cylin- 
der, the  cone,  and  the  sphere,  usually  referred  to  as  the  three 
round  bodies. 

CYLINDERS. 

566.  A  cylindrical  surface  is  a  curved  surface  generated 
by  a  straight  line  that  moves  so  as  continu-  ^,5 
ally  to  touch  a  given  curve,  while  remaining 
parallel  to  its  original  position.  Thus  if  the 
straight  line  Aa  moves  so  as  to  remain  always 
parallel  to  its  first  position  Aa,  while  contin- 
ually touching  the  curve  ABCD,  the  surface 
ABCB-ahcd  thus  generated  is  a  cylindrical 
surface.  The  moving  line  is  called  the  generatrix;  the 
curve  touched,  the  directrix;  and  any  straight  line  Bh,  that 
represents  the  generatrix  in  any  of  its  positions,  an  element 
of  the  surface. 

Since  the  generatrix  is  of  indefinite  length,  a  cylindrical 
surface  may  be  regarded  as  extending  indefinitely  in  two 
directions.  As  the  directrix,  moreover,  may  be  a  curve  of 
any  kind,  close  or  not,  the  surface  generated  may  present  a 
corresponding  variety  of  form.  In  elementary  geometry, 
for  obvious  reasons,  the  directrix  is  usually  assumed  to  be 

a  circle. 

293 


294  SOLID   GEOMETRY.  — BOOK  IX. 

567.  A  cylinder  is  a  solid  bounded  by  a  cylindrical  sur- 
face whose  directrix  is  a  closed  curve,  and  by  two  parallel 
planes.  These  planes  are  called  the  bases  of  the  cylinder ; 
the  curved  surface,  the  lateral  surface;  and  the  perpendicu- 
lar distance  between  the  bases,  the  altitude. 

From  the  definition  it  is  evident  that  the  curved  surface 
of  a  cylinder  must  have  as  directrix  a  closed  curve ;  since, 
otherwise,  besides  the  two  parallel  bases,  at  least  one  other 
plane  face  would  be  needed  in  order  to  form  a  solid. 

568.  A  right  cylinder  has  its  elements  perpendicular  to 
its  base,  as  AB ;  an  oblique  cylinder 

has  its  elements  oblique  to  its  base,       K~— - 
as  A'B\ 

569.  A  circular  cylinder  is  one 
that   has  a  circle  for  each   base. 

As    only    circular    cylinders    are    ^K' J      / / 

treated  of  in  this  book,  the  term  ^ 

cylinder  is  to  be  understood  as  signifying  circular  cylinder. 

570.  A  right  cylinder  with  a  circular  base  is 
called  a  cylinder  of  revolution,  because  it  may  be 
generated  by  the  revolution  of  a  rectangle  about 
one  of  its  sides  as  axis.  This  side  is  then  called 
the  axis  of  the  cylinder,  and  the  radius  of  the 
base,  the  radius  of  the  cylinder. 

Exercise  780.  Only  one  straight  line  can  be  drawn  through  a  given 
point  on  a  cylindrical  surface. 

781.  A  straight  line  that  joins  two  points  on  a  cylindrical  surface 
must  coincide  with  an  element  of  that  surface. 

782.  If  a  plane  contains  one  and  only  one  straight  line  in  com- 
mon with  the  curved  surface  of  a  cylinder,  the  plane  touches  but  does 
not  intersect  that  surface. 

Definitions.  Such  a  plane  is  said  to  be  tangent  to  the  cylinder, 
and  the  common  element  is  called  the  element  of  contact.  A  tangent 
line  touches  but  does  not  intersect  the  cylinder. 


•  CYLINDERS.  295 

Proposition  I.     Theorem. 

571.  Every  section  of  a  cylinder  made  by  a  plane 
passing  through  an  element  is  a  parallelogram. 


Given:  Ac,  a  section  of  cylinder  AB-c  by  a  plane  througli 
element  Aa; 

To  Prove :  Ac  is  a  parallelogram. 

For  if  througli  C  a  line  be  drawn  ||  to  Aa, 
that  line  will  be  an  element  of  the  surface;      (566) 
it  will  also  be  a  line  in  plane  Ac;  (424) 

,•.  the  line  will  coincide  with  (7c,  the  intersection 
of  plane  Ac  and  the  lateral  surface  ; 
.-.  Cc  is  II  to  Aa, 
and  ac  is  ||  to  AG,  (451) 

(since  they  lie  in  parallel  planes;) 

.*.  Ac  is  a  parallelogram.  q.e.d. 

« 
572.   Cor.  Every  section  of  a  right  cylinder  passing  through 
an  element  is  a  rectangle. 

Scholium.  It  will  be  noticed  that  the  properties  estab- 
lished in  Prop.  I.  and  Prop.  II.,  being  independent  of  the 
form  of  the  base,  hold  true,  not  only  in  regard  to  circular 
cylinders,  but  also  to  cylinders  in  general.  A  similar 
remark  applies  also  to  Prop.  III.  concerning  the  cone. 


296 


SOLID   GEOMETRY.  — BOOK  IX. 


Proposition  II.     Theorem. 
573.  The  bases  of  a  cylinder  are  equal. 


Given:  ABC,  dbc,  bases  of  cylinder  Ac; 

To  Prove :  ABC  equals  ahc. 

Through  any  element  Aa,  pass  planes  forming  the  sections 
Ah,  Ac;  and  join  BC,  he. 

Since  Ah  and  Ac  are  parallelograms,  (^^1) 

Bh  and  Cc  are  each  II  and  =  to  Aa;  (136) 

.-.  Ch  is  a  par'm,  and  BC  =  bc;  (142) 

.-.  AABC^Aahc.  (69) 

If,  then,  the  upper  base  be  applied  to  the  lower,  so  that 

ah  =^  AB, 

then  A  abc.^  A  ABC, 

and  c  ^  C; 

that  is,  any  point  c  in  the  perimeter  of  the  upper  base  will 
coincide  with  a  corresponding  point  in  the 
perimeter  of  the  lower  base ; 

.-.  the  bases  coincide  and  are  equal,     q.e.d. 

574.  Cor.  1.  Aiiy  two  parallel  sections,  s,  s', 
cutting  a  cylindncal  surface  MN,  are  equal. 

575.  Cor.  2.  All  sections  of  a  circular  cylin- 
der parallel  to  the  hase  are  equal. 


CONES.  297 

CONES. 

576.  A  conical  surface  is  a  curved  surface  generated  by 
a  straight  line  that  moves  so  as  continually  to  touch  a 
given  curve,  and  pass  through  a 
fixed  point  not  in  the  plane  of 
that  curve.  Thus  if  the  straight 
line  AA^  moves  so  as  always  to 
touch  the  curve  ABODE,  and  pass 
through  a  fixed  point  S,  the  sur- 
face S-ABCDE  is  a  conical  surface. 

The   moving   line    is   called   the  ,y//  \ 

generatrix;    the   curve   it   touches,  //// 

the  directrix;  the  fixed  point,  the     a'/"/"'T  e'    ■■■-., 

vertex.      Any  straight  line,  as  SB,     /  s^r;-, P' 

representing   one   of  the   positions 

of  the  generatrix,  is  an  element  of  the  conical  face. 

If  the  generatrix  is  of  indefinite  length,  the  whole  surface 
generated  consists  of  two  portions,  each  of  indefinite  extent, 
and  lying  on  opposite  sides  of  the  vertex,  the  one  being 
called  the  upper  nappe,  as  S-ABCDE;  the  other,  the  lower 
nappe,  as  S-A'b'c'd'e'. 

577.  A  cone  is  a  solid  bounded  by  a  conical  surface  and  a 
plane  cutting  that  surface,  as  S-ABCDE.  The  plane  sur- 
face is  called  the  base;  the  curved  surface,  the  lateral  sur- 
face; S,  the  vertex;  and  the  perpendicular  distance  from 
the  vertex  to  the  base,  the  altitude. 

578.  A  circular  cone  is  one  that  has  a  circu- 
lar base.  Its  axis  is  the  straight  line  drawn 
from  the  vertex  to  the  center  of  the  base. 

579.  A  right  circidar  cone  has  its  axis  perpen- 
dicular to  the  base ;  it  is  also  called  a  cone  of 
revolution,  because  it  may  be  regarded  as  gener- 
ated by  the  revolution  of  a  right  triangle  about 
one  of  its  arms  as  an  axis. 


298  SOLID   GEOMETRY.  — BOOK  IX. 

Proposition  III.     Theorem. 

580.  Every  section  of  a  cone  made  by  a  plane  pass- 
ing through  the  vertex  is  a  triangle. 


Given:  S-AC,  a  section  of  cone  s-ABD,  passing  through  ver- 
tex S  ; 

To  Prove :  S-AC  is  a  triangle. 

The  lines  SA,  sc,  are  elements  of  the  cone  ;     (576) 

they  also  lie  in  plane  SAC,  (9) 

(since  their  extremities  are  in  that  plane ;) 

.*.  SA,  SC,  are  the  intersections  of  the  plane  of  section  and 
the  lateral  surface. 

^C  is  also  a  straight' line,  (426) 

(being  the  intersection  of  two  planes ;) 

.-.  -S^C  is  a  triangle.  q.e.d. 

Exercise  783.  A  cylinder  of  revolution,  with  an  altitude  a  and  a 
radius  r,  rolls  on  a  plane  until  the  original  element  of  contact  again 
coincides  with  the  plane.  What  is  the  figure  of  the  plane  surface 
passed  over,  and  what  is  its  area  ? 

784.  A  cone  of  revolution,  altitude  a,  radius  r,  rolls  on  a  plane,  its 
vertex  remaining  fixed  until  the  original  element  again  coincides  with 
the  plane.    Name  the  figure  described,  and  give  its  area. 


CONES.  299 

Propositiox  IV.     Theorem. 

581.  Every  section  of  a  circular  cone  made  by  a 
plane  parallel  to  the  base  is  a  circle. 


Given :  A  section  abd  parallel  to  ABD,  the  base  of  circular 
cone  s-ABD  ; 

To  Prove :  abd  is  a  circle. 

Draw  SO,  the  axis  of  the  cone,  cutting  abd  in  o. 

Through  SO  and  any  elements,  SA,  SB,  etc.,  pass  planes 
cutting  the  base  in  the  radii  OA,  OB,  etc.,  and  abd  in 
oa,  ob,  etc. 

Since  OA  is  II  to  oa,  and  OB  is  II  to  ob,  etc.,      (451) 

(being  intersections  of  II  planes  by  a  third  plane,) 

A  Soa,  Sob,  etc.,  are  similar  to  A  SOA,  SOB,  etc.,  resp. ;  (292) 

.-.  oa:  0A=:  So:  so  =ob:  OB,  etc. 

But  0A  =  OB  ;  (162) 

.-.  oa  =  ob  =  oc,  etc. ; 

.',  section  abd  is  a  circle,        q.e.d.     (158) 

(since  all  st.  lines  drawn  from  o  to  its  perimeter  are  equal.) 

582.   Cor.    The  axis  of  a  circular  cone  passes  through  the 
center  of  all  sections  that  are  parallel  to  the  base. 


300  SOLID   GEOMETRY.  — BOOK  IX. 

SPHERES. 

583.  A  sphere  is  a  solid  bounded  by  a  curved  surface  all 
the  points  of  which  are  equally  distant  from  a  point  within, 
called  the  center.  Hence  a  sphere  may  be 
generated  by  the  revolution  of  a  semicir-  //a 
cle  about  its  diameter  as  axis.  //// 

584.  A  radius  of  a  sphere  is  any  straight   \  \  \  \ 
line  drawn  from  the  center  to  the  surface.     \N\ 
A   diameter    is   any   straight    line   drawn 
through  the  center  and  terminated  both  ways  by  the  surface. 

585.  CoR.  All  diameters  of  a  sphere  are  equal. 

Proposition  Y.     Theorem. 

586.  Every  section  of  a  sphere  made  by  a  plane  is 
a  circle. 


Given :  A  plane  section  ^^c  of  a  sphere  whose  center  is  0  ; 
To  Prove :  ABC  is  a.  circle. 

1°.  If  the  plane  passes  through  the  center  0,  then  the  lines 
drawn  from  0  to  any  points,  A,  B,  in  the  perimeter 
of  the  section,  are  equal,  being  radii  of  the  sphere ; 
therefore  ABC  is  a,  circle.  q.e.d.     (158) 

2°.  If  the  plane  does  not  pass  through  O,  draw  on  _L  to 
ABC,  and  join  0  and  D  with  A  and  B,  any  points  in 
the  perimeter  of  the  section. 


SPHERES.  301 

Since  radius  OA  =  radius  OB,  and  OD  is  common, 

rt.  A  ODA  =  rt.  A  ODB  (72),  and  DA  =  DB  ; 

.-.  ABC  18  d^  circle,  q.e.d.     (158) 

(since  any  two  points.  A,  B,  in  its  perimeter,  are  equally 
distant  from  D.) 

587.  Definition.  A  section  that  passes  through  the  cen- 
ter is  called  a  great  circle  of  the  sphere ;  a  section  not  pass- 
ing through  the  center,  a  small  circle  of  the  sphere. 

588.  Definition.  The  diameter  of  a  sphere  that  is  per- 
pendicular to  the  plane  of  a  circle  of  the  sphere  is  called 
the  axis  of  that  circle  ;  the  poles  of  the  circle  are  the  extremi- 
ties of  its  axis. 

589.  Cor.  1.  The  axis  of  a  circle  of  a  sphere  passes 
through  the  center  of  that  circle;  and  conversely. 

For  D,  the  foot  of  the  perpendicular  OB  from  the  center 
of  the  sphere,  is  the  center  of  the  circle  ABC. 

590.  CoR.  2.  A  circle  nearer  the  center  of  a  sphere  is 
greater  than  one  more  remote. 

For  the  less  the  distance  OD,  the  greater  is  DA,  the  radius 
of  the  circle. 

591.  CoR.  3.   All  great  circles  of  a  sphere  are  equal. 
For  their  radii  are  radii  of  the  sphere. 

592.  Cor.  4.  Any  two  great  circles, 
as  ABC,  abc,  bisect  each  other. 

For  since  they  have  the  same  cen- 
ter o,  their  intersection  Cc  is  a  diame- 
ter of  both  and  bisects  both. 

593.  Cor.  5.  Every  great  circle  bi- 
sects the  sphere. 

For  the  two  parts  into  which  the  sphere  is  divided  may 


302  SOLID   GEOMETRY.  — BOOK  IX. 

be  placed  so  as  to  coincide,  since  otherwise  there  would  be 
points  on  the  surface  unequally  distant  from  the  center. 

594.  CoR.  6.  An  arc  of  a  great  circle  may  he  made  to  pass 
through  any  two  given  points  A,  C,  on  the  surface  of  a  sphere. 

For  the  two  points,  A  and  C,  together  with  the  center  o, 
determine  (424)  the  plane  of  a  great  circle  whose  circum- 
ference passes  through  A  and  C. 

If  the  two  points,  as  A  and  B,  are  the  extremities  of  a 
diameter,  since  they  are  in  the  same  straight  line  with  the 
center  o,  any  number  of  planes  may  be  made  to  pass  through 
them  (421),  and  any  number  of  semicircumferences. 

595.  CoR.  7.  One  circle,  and  only  one,  can  pass  through 
three  given  points  on  the  surface  of  a  sphere. 

For  these  three  points  determine  one  plane  (424),  whose 
intersection  with  the  sphere  is  a  circle. 

Scholium.  Plane  loci,  as  we  have  seen  (211,  etc.),  are 
liyies  that  may  be  regarded  from  two  points  of  view,  either 
as  the  path  generated  by  one  point  moving  in  a  certain 
way,  or  as  the  assemblage  of  all  possible  points  having 
certain  properties  of  position.  Loci  in  space,  again,  are 
surfaces  that  may  be  regarded  either  as  generated  by  one 
line  moving  in  a  certain  way,  or  as  the  assemblage  of  all 
possible  lines  having  certain  properties  of  position.  Thus 
(435)  a  plane  may  be  regarded  as  generated  by  a  line  moving 
so  as  to  remain  always  perpendicular  to  a  given  line  at  a 
given  point.  Cylindrical  {^Q>Q>),  conical  (576),  and  spheri- 
cal surfaces  (583),  have  been  defined  as  generated  by  the 
motion  of  certain  lines  moving  in  a  certain  way.  From 
another  point  of  view,  the  plane  generated  as  above  men- 
tioned may  be  regarded  as  the  locus  of  all  perpendiculars 
to  the  given  line  at  the  given  point ;  a  circular  cylindrical 
surface,  as  the  locus  of  all  parallels  to  a  given  line  at  a 
given  distance ;  a  circular  conical  surface,  as  the  locus  of 
all  lines  inclined  to  a  given  line,  the  axis,  at  a  given  angle. 


SPHERES. 


303 


Proposition  VI.     Theorem. 

596.  All  points  in  the  circumference  of  a  circle  of 
a  sphere  are  equally  distant  from  each  pole. 


Given:  P,  P',  poles  of  a  circle  ABC  of  the  sphere  whose 
center  is  0; 

To  Prove:  All  points  in  the  circumference  ABC  are  equally 
distant  from  P  or  P'. 

Join  P  with  A,  B,  C,  any  points  in  circumf.  ABC. 

Since  OP  is  ±  to  circle  ABC,  .    (Hyp.) 

OP  passes  through  the  center  of  ABC ;         (589) 
.-.  PA  =  PB  =  PC,  (437) 

(being  obliques  from  P  to  points  A,  B,  C,  in  plane  ABC, 

that  are  equally  distant  from  the  ±  from  P;) 
i.e.,  all  points  in  circumference  ABC  are  equally  distant 
from  P.  Q.E.D. 

For  like  reasons  they  are  equally  distant  from  P'. 

597.  Cor.  1.  All  arcs  of  great  circles  draicn  from  a  pole 
of  a  circle  to  poiiits  in  its  circumference  are  equal. 

For  their  chords  are  equal  chords  (596)  of  equal  circles 
(591). 

598.  Definitions.  The  distance  between  two  given  points 
on  the  surface  of  a  sphere  is  the  arc  of  a  great  circle  join- 
ing those  points ;  the  polar  distance  of  a  point  A  in  the  cir- 


304  SOLID   GEOMETRY.  — BOOK  IX. 

cumference  of  a  circle  ^^c  is  the  arc   of  a  great  circle 
joining  A  and  the  nearer  pole  of  ^^  C 

599.  CoR.  2.  The  polar  distance  of  a  great  circle  a'b'c'  is 
a  quadrant;  that  is,  the  fourth  part  of  the  circumference  of  a 
great  circle. 

For  the  polar  distance  PmA^  measures  the  rt.  A  poa^ 
(262). 

600.  CoR.  S.  If  a  point  P  on  the  surface  of  a  sphere  is  at 
the  distance  of  a  quadrant  from  any  two  points,  A',  £',  in  an 
arc  of  a  great  circle,  then  P  is  the  pole  of  that  circle. 

For  the  arcs  PA\  PB',  being  quadrants,  the  angles  at  O 
are  right  angles ;  therefore  P  0  is  .L  to  OA'  and  to  OB' ;  hence 
it  is  ±  to  the  plane  of  arc  a'b'  (428);  whence  P  is  the  pole 
oisLTGA'B'  (588). 

601.  Scholium.  A  pole  of  a  circle,  great  or  small, 
whose  circumference  is  to  pass  through  a  given  point,  being 
known,  the  circumference  may  be  described  on  the  surface 
of  the  sphere. 

For   by   revolving  Pa,   an   arc   of  a         V^V^^^^^o 
great  circle,  about  the  pole  P,  the  ex-       /^^^--^.,y\ 
tremity  a  will  describe  the  small  circle    5^"   [  "Jo 

ahc;  while  the  extremity  of  PA,  a  quad-       y^^-Ld ^j 

rant,  will  describe  the  great  circle  ABC.  \^       ^/ 

Or,  placing   one   foot  of  the   spherical 
compasses*  at  P,  with  an  opening  between  the  feet  equal  to 
the  chord  of  the  polar  distance  of  the  point  a,  we  turn  the 
second  foot  round  the  sphere  so  as  to  describe  through  a 
the  required  circumference. 

Exercise  785.  Denoting  by  a  and  r  the  numerical  measures  of  the 
axis  and  radius  of  the  cylinder  rolling  as  in  Exercise  783,  what  must 
be  the  ratio  of  r  to  a  so  that  the  surface  generated  in  one  revolution 
shall  be  a  square  ? 

*  Compasses  with  the  feet  inclined  inwards. 


SPHERES. 


305 


Proposition  VII.    Problem. 

602.  To  find  the  radius  of  a  given  sphere, 
p 

....;p  /b 

^.""•••- !o 


Given :  A  sphere  APP' ; 

Required :  To  find  a  radius  of  APP\ 

Take  any  point  P  on  the  surface  as  pole,  and  with  any 
opening  of  the  compass  describe  a  circumference  AB  c  on  the 
sphere. 

Take  any  three  points,  A,  B,  C,  in  this  circumference,  and 
with  the  compasses  take  off  the  chord  distances  AB,  AC,  BO. 

On  any  plane  construct  the  triangle  A'b'c',  having  its  sides 
equal  to  AB,  AC,  BC,  respectively  (205),  and  circumscribe 
a  circumference  about  this  triangle  (185). 

This  circle  will  be  equal  to  circle  ABC,  and  its  radius  A'O 
will  be  equal  to  the  radius  of  circle  ABC. 

With  A'o  as  an  arm,  and  PA'  =  the  chord  of  the  arc  join- 
ing PA,  as  hypotenuse,  construct  the  rt.  A  PA'o. 

Draw  A'p'  ±  to  A'P,  and  produce  it  to  meet  PO  in  P'. 

It  is  evident  that  PP',  thus  determined,  is  equal  to  the 
diameter  of  the  sphere,  and  its  half,  PQ,  is  the  required 
radius. 

Exercise  786.  If  a  cone  of  revolution  roll  upon  a  plane,  its  vertex 
remaining  fixed,  what  kind  of  a  surface  is  generated  by  the  axis  of  the 
cone? 

787.   If  a  cone  of  revolution  roll  on  the  surface  of  a  second  cone,  so 
that  their  vertices  coincide,  what  kind  of  a  surface  is  generated  by  the 
axis  of  the  first  cone  ? 
Geom.  —  20 


306 


SOLID   GEOMETRY.  —  BOOK  IX. 


603.  Scholium  1.  The  radius  of  the  sphere  being  known, 
we  can  obtain  the  chord  of  a  quadrant  of 
that  sphere  by  finding  the  hypotenuse  h 
of  a  right  triangle  having  its  arms  each 
equal  to  r,  the  radius. 

604.  Scholium  2.  The  chord  of  a 
quadrant  of  a  sphere  being  found,  we  can 
describe  a  circumference  of  a  great  circle 
through  any  two  points.  A,  B,  on  the  sur- 
face of  the  sphere,  by  describing  from  A 
and  B  as  centers,  quadrants  intersecting 
in  P,  which  will  be  the  pole  of  the 
required  circumference  (600). 


Proposition  VIIT.     Theorem. 

605.  The  intersection  of  the  surfaces  of  two  spheres 
is  the  circumference  of  a  circle  perpendicular  to  the 
line  of  centers  of  the  spheres. 


Given :  00',  the  line  of  centers  of  two  intersecting  spheres  ; 
To  Prove :  The  intersection  of  the  surfaces  is  the  circumference 
of  a  circle  perpendicular  to  00'. 

Through  the  centers,  O,  0,'  let  a  plane  be  passed,  cutting 
the  two  spheres  in  great  circles  (587),  which  intersect  in 
A  and  B.  (Hyp.) 


SPHERES.  307 

Draw  the  chord  AB,  and  produce  00'  to  meet  the  circum- 
ferences. 

00'  is  X  to  ^5  at  its  mid  point  C.  (75) 

If  we  now  revolve  the  upper  part  of  the  figure  about  00', 
the  two  semicircles  will  generate  the  two  spheres  (583), 
while  the  point  A  will  generate  the  line  of  intersection  of 
the  surfaces. 

Also,  since  AC  during  the  revolution  remains  ±  to  00', 
AC  will  generate  a  circle  whose  center  is  C  ; 

i.e.,  the  intersection  of  the  surfaces  of  the  spheres  is  the 
circumference  of  a  circle  _L  to  00'.  q.e.d. 

606.  Definition.  Two  spheres  are  tangent  if  their  sur- 
faces have  but  one  common  point. 

607.  CoR.  If  tivo  spheres  are  tangent  to  each  other,  the 
point  of  tangency  is  in  their  line  of  centers. 

For  if  we  conceive  the  centers,  0,  O',  to  remove  farther 
from  each  other  till  the  circumferences  of  the  great  circles 
become  tangent,  the  points  A  and  B  will  coincide  with,  and 
the  circumference  of  intersection  be  reduced  to,  a  point  C. 

608.  Scholium.  Two  spheres  being  given  in  any  relative 
position,  a  plane  passed  through  their  centers  will  cut  them 
in  great  circles;  and  according  as  these  circles  are  within 
or  without,  are  tangent  or  intersecting,  the  spheres  will  have 
corresponding  positions  in  regard  to  each  other. 

609.  Definition.  A  plane  is  tangent  to  a  sphere  when  it 
has  but  one  point  in  common  with  the  surface  of  the  sphere. 

Exercise  788.  Denoting  by  a  and  r  the  altitude  and  radius  of 
a  cone  rolling  as  in  Ex.  784,  find  the  ratio  of  the  base  of  the  cone  to 
the  entire  circle  generated. 

789.  What  fraction  of  that  circle  is  described  by  one  revolution 
of  the  cone  ? 


308  SOLID   GEOMETRY.  — BOOK  IX. 

Proposition  IX.     Theorem. 

610.  A  plane  perpendicular  to  a  radius  of  a  sphere 
at  its  ejctremity  is  tangent  to  the  sphere. 


Given :  OP,  a  radius  of  a  sphere,  and  a  plane  MN  _L  to  OP  at 
its  extremity ; 

To  Prove :        MN  is  tangent  to  the  sphere. 

Take  any  point  except  P  in  MN,  as  A,  and  join  OA,  PA. 
Since  OP  is  ±  to  AP,  (Hyp.,  427) 

OA  >  op;  (99") 

.•.  A  lies  without  the  sphere  ; 
.-.  MN  is  tangent  to  the  sphere,    q.e.d.     (609) 
(since  every  point  in  MN,  except  P,  lies  without  the  sphere.) 

611.  Cor.  1.  A  plane  tangent  to  a  sphere  is  perpendicular 
to  the  radius  drawn  to  the  point  of  contact. 

612.  Cor.  2.  Any  straight  line  drawn  in  the  tangent  plane 
through  the  point  of  contact  is  tangent  to  the  sphere. 

613.  Cor.  3.  Any  two  straight  lines  tangent  to  the  sphere 
at  the  point  of  contact  determine  the  tangent  plane  at  that 
point. 


SPHERICAL  ANGLES  AND  POLYGONS. 


309 


SPHERICAL  ANGLES  AND  POLYGONS. 

614.  Definition.  The  angle  of  two  intersecting  curves 
is  the  angle  contained  by  the  two  tangents  to  the  curves  at 
the  common  point. 

This  definition  applies,  whatever  be  the  surface  upon 
which  the  curves  are  described. 

615.  Definition.  A  spherical  angle  is  the  angle  included 
between  two  arcs  of  great  circles  of  a  sphere.  The  arcs  are 
its  sides;  their  intersection  is  its  vertex. 


Proposition  X.     Theorem. 

616.  td  spherical  angle  is  measured  hy  the  arc  of  a 
great  circle  described  from  its  vertejc  as  pole,  and 
included  hy  its  sides,  produced  if  necessary. 


Given :  ^5,  an  arc  of  a  great  circle  described  from  the  vertex 
of  spherical  angle  APB  as  pole,  and  included  between  the  sides  of 
angle  APB  ; 

To  Prove :     Spherical  angle  APB  is  measured  by  mc  AB. 


Draw  PT,  pt',  tangents  to  PAP'^  pbp'  respectively,  and 
radii  OA,  OB. 


[]  (106) 


310  SOLID   GEOMETRY.  — BOOK  IX. 

Since  PT  is  ±  to  PP'  in  plane  PAP',      (Const.,  191) 
and  OA  is  ±  to  PP'  in  plane  PAP', 

(PA  being  a  quadrant,)  (Hyp.) 

PTis  II  to  0.4; 
similarly  PT'  is  II  to  OB 

.    ^  TPT'  =  Z  AOB.  (455) 

But  Z  AOB  i^  meas.  by  arc  ^5;  (262) 

le.,  spher.  Z  .IPi?  is  meas.  by  arc  AB.     q.e.d.     (614) 

617.  CoR.  1.  A  spherical  angle  has  the  same  measure  as 
the  dihedral  angle  formed  by  the  planes  of  its  sides. 

618.  Cor.  2.  All  arcs  of  great  circles  drawn  through  the 
pole  of  a  given  great  circle  are  perpendicular' to  its  circumfer- 
ence. 

For  their  planes  are  each  perpendicular  to  its  plane  (469). 

619.  Scholium.  The  foregoing  corollary  enables  us, 
through  any  giv^en  point  P  on  the 
surface  of  a  sphere,  to  describe  an 
arc  of  a  great  circle  perpendicular  to 
a  given  arc  ABC  of  a  great  circle. 
From  P  as  pole  describe  (428)  an 
arc  of  a  great  circle  cutting  ABC  in 
C,  and  from  C  as  pole  describe  an 
arc  of  a  great  circle  passing  through 
P  and  cutting  ABC  in  B  ;  then  arc  PB  is  _L  to  arc  ABC  (618). 

620.  Definition.  A  spherical  polygon  is  a  portion  of  the 
surface  of  a  sphere  bounded  by  three  or  more  arcs  of  great 
circles.  The  bounding  arcs  are  the  sides  of  the  polygon; 
the  angles  formed  by  these  sides  are  the  angles,  and  the 
vertices  of  the  angles  are  the  vertices,  of  the  polygon. 


SPHERICAL  ANGLES  AND  POLYGONS.         311 

621.  Since  the  planes  of  all  great  circles  pass  through 
the  center  of  the  sphere,  the  planes  of  the  sides  of  a  spheri- 
cal polygon  form  at  the  center  a  polyhedral  angle  whose 
edges  are  radii  drawn  to  the  vertices  of  the  polygon ;  the 
face  angles  of  the  polyhedral  angle  are  angles  at  the  center 
measured  by  the  arcs  that  form  the  sides  of  the  polygon ; 
while  the  dihedral  angles  of  the  polyhedral  angle  have  the 
same  measure  as  the  angles  of  the  polygon  (616). 

Thus  the  planes  of  the  sides  of  the  polygon  An  CD  -{see 
diagram  for  Prop.  XII.)  form  at  0,  the  center  of  the  sphere, 
the  polyhedral  angle  0-ABCD.  The  face  angles,  AOB,  BOC, 
etc.,  are  measured  by  the  sides  AB,  BC,  etc.,  of  the  poly- 
gon; and  the  dihedral  angle  whose  edge  is  the  radius  OA, 
has  the  same  measure  as  the  spherical  angle  BAD,  etc. 

622.  From  the  relations  thus  established  between  poly- 
hedral angles  and  spherical  polygons,  it  clearly  follows 
that  from  any  known  property  of  polyhedral  angles  we  may 
infer  a  corresponding  property  of  spherical  polygons;  and  con- 
versely. 

623.  Definition.  A  diagonal  of  a  spherical  polygon  is 
an  arc  of  a  great  circle  passing  through  two  nonadjacent 
vertices. 

624.  Definition.  A  spherical  triangle  is  a  spherical  poly- 
gon having  three  sides.  Like  plane  triangles,  spherical  tri- 
angles may  be  right  or  oblique,  scalene,  isosceles,  or  equilateral. 


Exercise  790.  Out  of  a  circle  with  radius  B  a  sector  of  60°  is  cut, 
and  the  edges  of  the  remaining  sector  are  joined  so  as  to  form  the 
lateral  surface  of  a  cone  of  revolution.  Find  the  radius  r,  and  the 
altitude  a,  of  the  cone  thus  formed. 

791.  Find  general  formulas  for  the  radius  r  and  the  altitude  a  of 
a  cone  of  revolution  whose  lateral  surface  is  formed  from  a  sector 

that  is  ~i\\^  of  a  circle  with  radius  B. 


312 


SOLID   GEOMETRY.  —  BOOK  IX. 


Proposition  XI.     Theorem. 

625.  Any  side  of  a  spherical  triangle  is  less  than 
the  suin  of  the  other  two. 

c 


■...,A 


O-.;::;- 


Given :  A  spherical  triangle  ^^c,  on  a  sphere  whose  center  is  o  ; 
To  Prove :       AB  -\-  AC  is  greater  than  BC. 

In  the  trihedral  angle  0-ABC, 
Zaob-{-Zaoc>Zboc;  (491) 

.-.  arc^5 +arc^C>arc5(7.     q.e.d.     (616) 


626.  Cor.  1.  Any  side  of  a  spherical  triangle  is  greater 
than  the  difference  of  the  other  two. 

627.  Cor.  2.  The  shortest  path  07i  the  surface  of  a  sphere 
between  two  given  points,  A  and  B,  is  the  arc  A  MB  of  a  great 
circle  passing  through  those  points. 

For  let  ACB  be  any  other  curve  join- 
ing A  and  B.  Take  in  it  any  point  C,  and 
through  A,  G,  and  C,  B,  describe  arcs  of 
great  circles  (594).  Then  AMB  <AC  +  CB 
(625).  Between  A  and  C,  C  and  B,  take 
any  points,  D  and  E,  and  describe  arcs  of 
great  circles  joining  AD,  DC,  CE,  EB ;  then 
AC  -\-  CB  <  AD  -\-  DC  -\-  CE  -[■  EB.  If  this 
process  be  continued  indefinitely,  the  sum  of  the  arcs  thus 
obtained  will  be  always  increasing,  and  approaching  the 
curve  ACB  as  its  limit;  hence  J C^  must  be  greater  than 
AMB. 


POLAR    TRIANGLES. 


313 


Proposition  XII.     Theorem. 

628.  The  sum  of  the  sides  of  any  spherical  polygon 
is  less  than  the  circumference  of  a  great  circle. 


0-: 


Given:  A  spherical  polygon  ^J5CI>,  on  a  sphere  with  center  0 ; 
To  Prove :  AB-\-BC-{-CD-{-DA  <d,  circumf.  of  a  great  circle. 

Draw  the  radii  OA,  OB,  OC,  OD. 
In  the  polyhedral  angle  0-ABCD, 
ZAOB  -^ZBOC-i-ZcOB  -{-ZdOA<  four  rt.  A;    (492) 
arc  AB-\-SiicBC+SiTG  CZ>  H-arc  i)^  <a  circumf.  of  a  great O. 

Q.E.D.      (616) 


POLAR  TRIANGLES. 

629.  Definition.  If  from  the  vertices  of  a  spherical 
triangle  as  poles,  arcs  of  great  circles  be  described,  they 
will  form  by  their  intersection  a 
second  triangle,  called  the  polar 
triangle  of  the  first.  Thus  if  A, 
B,  C,  the  vertices  of  the  spherical 
triangle  ABC,  are  the  poles  of  the 
arcs  b'c',  A'C',  a^b\  forming  the 
spherical  triangle  A^B'c\  then 
A'b^C^  is  the  polar  triangle  of  ABC. 

Since  all  circumferences  of  great  circles  intersect  in  two 


314 


SOLID   GEOMETRY.— BOOK  IX. 


points,  the  arcs  A^B\  A'c',  B'c',  if  produced,  will  form  three* 
other  triangles  on  the  surface  of  the  sphere.  Hence  it  is 
to  be  understood  that  the  triangle  to  be  taken  as  the  polar 
triangle  oi  AB  c  is  the  central  one,  whose  vertex  A',  homol- 
ogous to  A,  is  on  the  same  side  of  -B  c  as  the  vertex  A,  and 
similarly  in  regard  to  the  other  vertices. 


Proposition  XIII.     Theorem. 

630.  If  the  first  of  two  spherical  triangles  is  the 
polar  triangle  of  the  second,  reciprocally,  the  second 
triangle  is  the  polar  triangle  of  the  first. 


Given :  A'b'c',  the  polar  triangle  of  ABC; 

To  Prove :    ABC  \^  the  polar  triangle  of  A' b'c'. 

Since  5  is  a  pole  of  A'c',  an  arc  of  a  great  O,    (Hyp.) 
the  distance  of  B  from  A'  is  a  quadrant.  (599) 

Since  C  is  a  pole  of  A'b',  an  arc  of  a  great  O, 
the  distance  of  C  from  A'  is  a  quadrant ; 

.*.  A'  is  a  pole  of  B  c,  (600) 

(being  at  a  quadrant's  distance  from  its  extremities. ) 
Similarly,  B'  and  C'  are  poles  of  ^  C  and  A  B  resp. 
Also,  a'  and  A  are  on  the  same  side  of  B  C,  etc. ; 
.-.  ^i?c  is  the  polar  triangle  of  a'b'c'.     q.e.d.    (629) 

*  Excluding  those  having  sides  greater  than  a  semicircumference. 


POLAR    TRIANGLES.  315 

Proposition  XIV.     Theorem. 

631.  In  two  polar  triangles,  each  angle  of  the  one 
is  measured  hy  the  supplement  of  the  opposite  side 
of  the  other. 


Given :       ABC  and  A'b'c',  two  polar  triangles  ; 

To  Prove :  Angle  A  is  measured  by  180°  —  arc  b'c',  etc. 

Produce  AB,  AC,  if  necessary,  to  meet  B'c'  in  D,  E  resp. 
Since  A  is  the  pole  of  arc  DE,  (Hyp.) 

Z  ^  is  measured  by  arc  DE.  (616) 

Since  5'  and  C'  are  poles  of  arcs  AE,  AD,  resp.,  (Hyp.) 
arcs  B'E  and  C^D  are  each  quadrants ; 
.-.  B^E  +  C\d  =  a  semicircumf. ; 
.-.  B'c'  +  D^  =  a  semicircumf.,  =  180° ; 
.-.  DE,  which  measures  Z  A,  =  180°  —  B'c'. 
In  the  same  way  it  may  be  proved  that  A  B  and  C  are 
measured  by  the  supplements  of  arcs  A'c'  and  a'b'  resp.  q.e.d. 

632.  Definition.  Polar  triangles  are  also  called  sup- 
plemental triangles.  For  if  we  denote  by  a'  the  number  of 
degrees  in  B^c',  the  side  of  A' B'c'  that  is  opposite  Z  A,  etc., 
we  have  the  relations : 

•  Za  =  180°  -  a',     Zb  =  180°  -  h',     Z  C  =180°  -  c', 
Z  A'  =  180°  -a,     Zb'  =  180°  -h,      Z  c'  =  180°  -  c. 


316  SOLID   GEOMETRY.  — BOOK  IX. 

Proposition  XV.     Theorem. 

633.  The  sum  of  the  angles  of  a  spherical  triangle 
is  greater  than  two,  and  less  than  six,  right  angles. 


O- 


K^    /. 


Given :     A,  B,  c,  the  angles  of  a  spherical  triangle  ABC; 
To  Prove :       Aa-\-Z.b-\-Z.c>  180°  and  <  540°. 

Construct  A'b'c',  the  polar  triangle  of  ABC.  Then  denot- 
ing (as  in  Art.  632)  the  number  of  degrees  in  B'c', 
A'C',  A'b',  resp.  by  a',  b',  c', 

since  Za  =  180° -a',  Zb  =  180° -6',  Z  c  =180°-  c',  (632) 

Z  A-]-Zb  -{-Z  c  =  540°  -  (a'  +  6'  +  c').      (Ax.  2) 

But  a'  +  b'-hc'  <  360°  and  >  0°;  (492) 

.'.Za+Zb  +  Zc>  (540°  -  360°)  or  180°, 

andZ  A  +  Z  B  -\-  Z  c  <  (540°  -  0°)  or  540°.     q.e.d. 

634.  Cor.  A  spherical  triangle  may  have  two,  or  even 
three,  right  angles;  also  two,  or  even  three,  obtuse  angles. 

635.  Definition.  A  spherical  triangle  having  two  right 
angles  is  said  to  be  birectangular ;  a  spherical  triangle  hav- 
ing three  right  angles  is  said  to  be  trirectangular. 

636.  Definition.  The  excess  of  the  sum  of  the  angles 
of  a  spherical  triangle  over  two  right  angles  is  called  fhe 
spherical  excess  of  the  triangle. 


POLAR    TRIANGLES. 


317 


Thus,  denoting  its  spherical  excess  by  E,  we  have  for 
A  ABC, 

e=Za-^Zb-\-Zc-  180°. 

637.  Definition.  The  spherical  excess  of  any  spherical 
polygon  of  n  sides  is  equal  to  the  excess  of  the  sum  of  its 
angles  over  2{n  —  2)  right  angles;  that  is,  is  equal  to  the 
sum  of  the  spherical  excess  of  the  {n  —  2)  spherical  tri- 
angles into  which  the  polygon  can  be  divided  by  means 
of  arcs  drawn  from  any  vertex  to  the  opposite  vertices. 

638.  Definition.      Symmetrical    spherical   triangles  are 
such  as  have  their  sides  severally  equal,  but 
arranged  in  reverse  order;  as  ABC,  A'b'c'. 

A'b'c'  may  be  regarded  as  formed  by  pro- 
ducing AO,  BO,  CO,  to  meet  the  surface  of 
the  sphere  in  A',  B',  c',  and  joining  the 
points  thus  obtained  by  arcs  of  great  circles. 
We  can  conceive  A'b'c',  when  thus  formed, 
as  moved  to  any  other  position  on  the  spheri- 
cal surface;  and  vice  versa. 

Since  the  face  angles  of  the  trihedrals 
0-ABC,  0-A'b'c',  are  equal,  but  arranged 
in  reverse  order,  the  trihedrals  are  symmetrical  (493),  and 
cannot  be  made  to  coincide  unless  isosceles  (497).  Hence 
the  spherical  triangles  whose  sides  are  the  intersections  of 
the  planes  of  the  faces  of  the  trihedrals,  cannot  be  made  to 
coincide  unless  these  trihedrals  can  be  made  to  coincide ; 
that  is,  unless  they  are  isosceles. 

639.  CoR.  An  isosceles  spherical  triangle  can  he  made  to 
coincide  with  its  symmetrical  triangle. 

For  as  the  trihedral  angles  formed  by  the  planes  of  their 
sides  can  be  made  to  coincide  (497),  the  sides  formed  by 
the  intersections  of  those  planes  with  the  surface  can  be 
made  to  coincide. 


318  SOLID   GEOMETRY.  — BOOK  IX. 

Proposition  XVI.     Theorem. 

640.  Two  triangles  on  the  saine  sphere  are  either 
equal  or  syminetrical,  if  a  side  and  the  including 
angles  of  the  one  are  respectively  equal  to  a  side  and 
the  including  angles  of  the  other. 


Given:  In  spherical  triangles  ABC,  a'b'c',  ab  equal  to  A^B\ 
angle  A  equal  to  angle  A\  and  angle  B  equal  to  angle  B' ; 
To  Prove:  ABC  and  A'b'c'  are  either  equal  or  symmetrical. 

For  A  ABC  may  be  placed  either  upon  A  A'b'c'  or  upon 
A  abc,  symmetrical  with  A'b'c',  so  as  to  coincide,  as  may  be 
shown  by  the  same  course  of  reasoning  as  that  employed  in 
Prop.  VI.,  Book  I.  Hence  A  ^J5C  is  either  equal  to  A  a'b'c' 
or  symmetrical  with  it.     q.e.d. 

641.  Cor.  If  a  spherical  triangle  has  two  equal  angles,  it  is 
isosceles. 

For  ii  Z  A  =  Z  b  =  Z  A'  =  Z  b',  then,  also,  since  Z  A—Z  a 
and  Zb  ^  Zh,  A  ABC  can  be  made  to  coincide  with  both 
A  a'b'c'  and  with  its  symmetrical  triangle  abc.  Hence 
a'b'c'  and  ahc  must  be  isosceles  (639);  whence  also  ABC  is 
isosceles. 

Exercise  792.  If  two  sides  of  a  spherical  triangle  are  quadrants, 
the  third  side  measures  the  opposite  angle. 

793.  A  spherical  triangle  ABC  has  Z^  =  83o,  /  ^  =  50°,  and 
AC  —  97°  ;  find  how  many  degrees  there  are  in  the  sides  of  the  polar 
triangle  that  are  respectively  opposite  to  those  angles. 


POLAR    TRIANGLES.  319 

Proposition  XVII.     Theorem. 

642.  Two  triangles  on  the  same  sphere  are  either 
equal  or  symmetrical,  if  two  sides  and  the  included 
angle  of  the  one  are  respectively  equal  to  two  sides 
and  the  included  angle  of  the  otJier. 


Given:  In  spherical  triangles  ABC,  a'b'c',  AB  equal  to  A^b\ 
AC  equal  to  A'c\  and  angle  A  equal  to  angle  A^ ; 

To  Prove :  ABC  and  a'b'c'  are  either  equal  or  symmetrical. 

For  A  ABC  can  be  made  to  coincide  either  with  A  A'b'c', 
or  with  its  symmetrical  triangle  abc,  as  may  be  shown  by 
the  same  course  of  reasoning  as  that  employed  in  Prop. 
VIII.,  Book  I.  Hence  A  ABC  is  either  equal  to  Aa'b'c' 
or  symmetrical  with  it.     q.e.d. 

643.  Cor.  1.  In  an  isosceles  spherical  triangle,  the  angles 
opposite  the  equal  sides  are  equal. 

For  if  ABC  is  isosceles,  then  both  A'b'c'  and  its  sym- 
metrical triangle  abc  must  be  isosceles.  Hence,  as  Z  B 
can  be  made  to  coincide  with  Z  c,  and  Z  c  =  Z  c'  =  Zc, 
Zb  =  Zc. 

644.  Cor.  2.  The  arc  of  a  great  circle  drawn  from  the 
vertex  of  an  isosceles  spherical  triangle  to  the  mid  point  of  the 
base,  bisects  the  vertical  angle,  is  perpendicular  to  the  base,  and 
divides  the  triangle  into  two  symmetrical  triangles. 

Exercise  794.  If  a  spherical  triangle  ABC  is  isosceles,  its  polar 
triangle  A'B'C  is  also  isosceles. 


320  SOLID  GEOMETRY.  — BOOK  IX. 

Proposition  XVITI.     Theorem. 

645.  Two  triangles  on  the  same  sphere  are  either 
equal  or  symmetrical,  if  the  three  sides  of  the  one  are 
respectively  equal  to  the  three  sides  of  the  other. 

For  as  the  respectively  equal  sides  must  be  arranged 
either  in  the  same  or  in  the  reverse  order,  the  one  triangle 
can  be  made  to  coincide  either  with  the  other  or  with  its 
symmetrical  triangle,  as  may  be  shown  by  a  course  of  reason- 
ing similar  to  that  employed  in  Prop.  X.,  Book  I.     q.e.d. 

646.  Scholium.  This  and  other  theorems  in  regard  to 
triangles,  etc.,  on  the  same  sphere,  will  evidently  hold  true  in 
regard  to  triangles,  etc.,  on  equal  spheres ;  since,  if  the  cen- 
ters of  the  equal  spheres  be  made  to  coincide,  their  surfaces 
will  also  coincide. 


Proposition  XIX.     Theorem. 

647.  If  two  triangles  on  the  same  sphere  are  mu- 
tually equiangular,  they  are  mutually  equilateral 
and  are  either  equal  or  syxnmetrical. 


Given :  Two  mutually  equiangular  spherical  triangles  A  and  A^ ; 

To  Prove :  A  and  A^  are  mutually  equilateral  and  either  equal 
or  symmetrical. 


POLAR    TRIANGLES.  321 

Construct  P,  P',  the  polar  triangles  of  A,  A'  respectively. 

Since  A  and  A'  are  mutually  equiangular,         (Hyp.) 

P  and  P',  their  polar  A,  are  mutually  equilateral,     (631) 

(their  corresponding  angles  being  measures  of  equal  angles ;) 

.-.  P  and  P'  are  mutually  equiangular;  (645) 

.-.  A  and  A',  the  polar  A  of  P  and  P',  are  mutually  equi- 
lateral; (631) 

.-.  A  and  A'  are  equal  or  symmetrical,  q.e.d.  (645) 

Scholium.  Mutually  equiangular  spherical  triangles  are 
equilateral  only  when  on  the  same  sphere  or  equal  spheres. 
When  the  equiangular  triangles  are  on  unequal  spheres, 
their  homologous  sides  are  proportional  to  the  radii  of  their 
spheres ;  the  triangles  are  then  said 
to  be  similar. 

648.  CoR.  If  three  planes  are 
passed  through  the  center  of  a  sphere, 
each  perpendicular  to  the  other  two, 
they  divide  the  surface  into  eight 
equal  trirectangular  triangles  (621, 
647). 

Exercise  795.   Birectangular  triangles  on  equal  spheres  are  equal 
if  their  acute  angles  are  equal. 

796.  Birectangular  triangles  on  unequal  spheres  are  similar  if  their 
acute  angles  are  equal. 

797.  Trirectangular  triangles  on  equal  spheres  are  equal. 

798.  Trirectangular  triangles  on  unequal  spheres  are  similar. 

799.  The  polar  triangle  of  a  birectangular  triangle  is  birectangular. 

800.  In  a  birectangular  triangle,  the  sides  that  are  opposite  to  the 
right  angles  are  quadrants. 

801.  Each  side  of  a  trirectangular  triangle  is  a  quadrant. 

802.  The  polar  triangle  of  a  trirectangular  triangle  is  a  trirect- 
angular triangle  coinciding  with  the  triangle  itself. 

Geom.  — 21 


322  SOLID   GEOMETRY.  — BOOK  IX. 

Proposition  XX.     Theorem. 

649.  Symmetrical    spherical    triangles    are  equiv- 
alent. 

.■■:/tA  y^     . 


p:. 


Given:  In  symmetrical  triangles  ABC,  A'b'c',  AB  equal  to 
A's',  AC  equal  to  A'c',  BC  equal  to  £'c'; 

To  Prove:  Triangle  ABC  is  equivalent  to  triangle  A'b'c'. 

Find  P  and  P',  the  poles  of  small  circles  passing  through 
A,  B,  c,  and  A',  b'j  c'  respectively. 

Since  arcs  AB,  AC,  5C  =  arcs  a'b',  A'c',  b'c'  resp.,    (Hyp.) 
the  chords  of  arcs  AB,  AC,BC  =  the  chords  of  arcs  A'b',  A'c', 

b'c'  resp. ; 
.*.  the  plane  triangles  formed  by  these  chords  are  equal; 

(69) 

.-.  the  small  circles  through  A,  B,  C,  and  A',  B',  c',  are  equal. 

Through  PA,  PB,  PC,  P'A',  p'b',  p'c',  pass  arcs  of  great 

circles. 

Since  arcs  PA,  PB,  PC,  P'A',  p'b',  p'c',  are  equal,    (596) 

(being  polar  distances  of  equal  circles  on  the  same  sphere,) 

A  PAB,  p'a'b',  are  symmetrical  and  isosceles; 

.-.  A  PAB  =  A  P'A'B'.  (639) 

Similarly,  A  PBC  =  A  p'b'c',  and  A  PAC  =  A  P'A'c' ; 
.'.  PAC  -{-PBC—  PAB  ^  P'A'c'  -\-  p'b'c'  —  p'a'b' ; 

(Ax.  2,  Ax.  3.) 
.-.  A  ABC  =0^  A  a'b'c'.  Q.e.d. 


POLAR    TRIANGLES.  323 

If  pole  P  should  lie  within  A  ABC,  then  pole  P'  would 
also  lie  within  Aa'b'C',  and  we  should  have 

PAB+  PAC-}-  PBC^P'A'b'  -{-  P'A'C'-\-P'B'C'; 

.'.  A  ABC^  AA'B'C'. 


Proposition  XXI.     Theorem. 

650.  In  a  spherical  triangle,  a  greater  side  is  oppo- 
site a  greater  angle. 


A 


Given:  In  spherical  triangle  ABC,  angle  A  greater  than 
angle  B; 

To  Prove :  i?  C  is  greater  than  A  c. 

Through  A  describe  AD,  an  arc  of  a  great  O,  so  that 
Zbad  =  Zb. 

Bince  Zbad=  Zb,  BD=DA;  (641) 

.'.   BD -\-DC=DA-\- DC; 
^  .:   BC>  AC.  Q.E.D      (625) 

651.  Cor.  Conversely,  if,  in  triangle  ABC,  BC  is  greater 
than  AC,  then  angle  A  is  greater  than  angle  B. 

For  these  angles  cannot  be  equal,  since  BC  and  AC  are 
not  equal  (Hyp.). 

Nor  can  Z  Ahe  less  than  Z  B,  for  then  BC  would  be  less 
than  AC  (650)  j  hence  Z  A  must  be  greater  than  Z  B. 


324 


SOLID   GEOMETRY.  — BOOK  IX. 


652.  Definition.    A  lu7ie  is  the  part  of  the  surface  of  a 
sphere  that  is  included  between  two  semi- 
circumferences  of  great  circles;  as  AMBN.' 

The  angle  of  the  lune  is  that  between  the 
semicircumferences  that  form  its  sides. 

653.  Cor.    Lunes  on  the  same  sphere  hav- 
ing equal  angles  are  equal. 

For  they  evidently  can  be  made  to  coincide. 

654.  Definition.  A  spherical  ungula  or  wedge  is  the  part 
of  a  sphere  bounded  by  a  lune  as  base,  and  by  the  planes  of 
its  sides ;  as  AON  MB. 

The  angle  of  the  ungula  is  the  same  as  the  angle  of  its 
base ;  the  diameter  AOB  is  called  the  edge  of  the  ungula. 


Proposition  XXIL     Theorem. 

655.  If  two  arcs  of  great  circles  intersect  on  the 
surface  of  a  hemisphere,  the  sum  of  the  two  opposite 
triangles  thus  formed  is  equivalent  to  a  lujie  whose 
angle  is  equal  to  that  formed  hy  the  arcs. 


Given:  On  the  hemisphere  A-BDCE,  two  arcs  of  great  circles, 
BAC  and  DAE,  intersecting  at  A,  and  forming  the  A  ABD,  ACE ; 
To  Prove :  A  ABD  -i- A  ACE  ^  el  lune  whose  Z  =  Z  CAE. 


POLAR    TRIANGLES.  325 

Produce  arcs  AC,  AE,  till  they  meet  at  A'. 

Since  BAC  =  ^  semicircumf.  =  ACA',         (Hyp.) 

BAG  —  AC=  ACA' —  AC;  (Ax.  3) 

i.e.,  AB  =A'C. 

In  the  same  way  it  may  be  shown  that 

AD  =  a'e  and  BD  =  EC; 

.-.  Aabd,  a'ce,  are  mutually  equilateral; 

.-.  AABD  ^  A  A'ce;  (645,  649) 

.-.  AABD  -}- A  ACE  -0=  A  A'CE  +  A  ACE; 
i.e.,  A  ABD  -[-AACE<^  lune  ACA'e, 

whose  Z  is  Z  CAE.  q.e.d. 

656.  Definition.  Just  as  the  angle  which  is  the  nine- 
tieth part  of  a  right  angle  is  called  a  degree,  so  the 
birectangular  triangle  which  is  the  ninetieth  part  of  a 
trirectangular  triangle,  and  has  for  base  the  arc  that  meas- 
ures one  degree,  is  called  a  spherical  degree.  Denoting  by 
T  the  surface  of  a  trirectangular  triangle,  the  surface  of  the 
sphere  =  8  T  =  720  spherical  degrees.  It  is  also  obvious 
that  the  lune  Avhose  angle  is  one  degree  will  contain  two 
spherical  degrees. 

It  is  important  to  keep  in  mind  the  distinction  between  the  three 
different  senses  in  which  the  term  degree  is  employed.  A  degree  of 
angular  measure  is  the  90th  part  of  a  right  angle,  a  relative  position  of 
two  straight  lines ;  an  arc  degree  is  a  line,  the  90th  part  of  a  quadrant 
or  360th  part  of  a  circumference  ;  a  spherical  degree  is  a  surface,  the 
90th  part  of  a  trirectangular  triangle  or  the  360th  part  of  the  surface 
of  a  hemisphere. 

Exercise  803.  If  a  spherical  triangle  has  one  right  angle,  the  sum 
of  the  acute  angles  is  greater  than  a  right  angle. 

804.   Lunes  with  equal  angles  on  unequal  spheres  are  similar. 


326 


SOLID   GEOMETRY.  — BOOK  IX. 


Proposition  XXIII.     Theorem. 

657.  A  lune  is  to  the  surface  of  the  sphere  as  the 
angle  of  the  lune  is  to  four  right  angles. 


Given:  A  lune  L  upon  a  sphere  whose  surface  is  S,  and  AOB, 
the  angle  of  the  lune,  whose  poles  are  P,  P' ; 
To  Prove :  L:  S  =  angle  AOB  :4:  right  angles. 

1°.  When  arc  AB  and  circumference  ABC  are  commensu- 
rable. 

Let  arc  AE  be  a  common  measure  of  arc  AB  and  the  cir- 
cumference, so  that  AE  can  be  laid  off  9  times  on  AB  and 
73  times  on  ABC. 

Suppose  arcs  of  great  circles  to  be  passed  through  the 
points  of  division  and  the  poles  P,  P'.  The  lunes  whose 
angles  are  measured  by  the  equal  arcs  are  equal  (653),  and 
each  is  contained  9  times  in  L  and  73  times  in  S. 

Since  Z  ^  05  : 4  rt.  Zs  =  9  :  73,  (Hyp. ) 

and  i:S  =  9:73,  (Const.) 

L:S  =  ZA0B:4:Vt.  A.       Q.E.D.     (232'") 

2°.  When  arc  AB  and  the  circumference  are  incommensu- 
rable, we  can  prove  by  the  method  of  limits  (as  in  Prop. 
II.,  Book  IV.)  that  always 

L:S  =  Zaob  :  4:  rt  A.  q.e.d. 


POLAR    TRIANGLES.  327 

658.  Cor.  1.  On  the  same  sphere,  lunes  are  to  each  other 
as  their  ayigles. 

659.  Cor.  2.     Denoting  by  A  the  number  of  degrees  in  the 

angle  of  a  lune, 

since  X  :  8  r  =  ^  :  360, 

45 

660.  Cor.  3.  The  spherical  ungula  AB  —  PP'  :  sphere 
=  ZaoB:S60. 

For  ungulas  are  equal  if  their  lunes  are  equal,  since  they^ 
can  be  made  to  coincide.     Hence  in  equal  spheres,  ungulas 
are  to  each  other  as  their  lunes;  or,  employing  V  to  denote 
the  volume  of  the  sphere,  and  u  that  of  the  ungula,  we  have 

U:  V  =  A:360; 

45 

661.  Scholium.  In  the  above  formulas,  the  symbols 
L,  T,  U,  V,  denote  only  relative  values.  In  order  to  obtain 
the  absolute  value  of  L  or  U,  we  must  know  that  of  T  or  V, 
which,  again,  depend  upon  the  radius  of  the  sphere;  in 
what  manner  will  be  seen  further  on. 

Exercise  805.  The  angle  of  a  lune  is  36°.  What  fraction  is  the 
lune  of  the  surface  of  its  sphere  ? 

806.  A  lune  comprises  one  hundredth  part  of  the  surface  of  a 
sphere.    What  is  the  angle  of  that  lune  ? 

807.  A  trirectangular  triangle  is  to  a  lune  on  the  same  sphere  as  12 
is  to  5.  What  is  the  angle  of  the  lune,  and  what  fraction  is  it  of  the 
whole  spherical  surface  ? 

808.  The  dihedral  angle  formed  by  the  plane  faces  of  a  spherical 
wedge  is  25°.     What  fraction  is  that  wedge  of  the  whole  sphere  ? 

809.  In  order  that  a  spherical  wedge  shall  be  J^th  of  its  sphere, 
what  must  be  the  angle  of  its  base  ? 


328  SOLID   GEOMETRY.  — BOOK  IX. 

Proposition  XXIV.     Theorem. 

662.  A  spherical  triangle  is  equivalent  to  as  many 
spherical  degrees  as  there  are  angular  degrees  in  its 
spherical  excess. 


Given :  A  spherical  A  AB  c,  whose  spherical  excess  is  E  degrees ; 
To  Prove :  Triangle  AB C  is  equivalent  to  E  spherical  degrees. 

Produce  the  sides  to  meet  DKH,  a  great  circle  described 
about  ABC. 
Since  A  ADF  +  A  AEG  <^  a  lune  whose  Z=  Za,  (655) 
A  ADF  -{-A  AEG  ^2  A  spherical  degrees.     (656) 
In  the  same  way  it  may  be  proved  that 
A  BEK  +  A  BDH <^2  B  spher.  deg., 
A  CGH  +  A  CFK  =0=  2  (7  spher.  deg. 
Now  the  sum  of  these  six  triangles  exceeds  the  surface  of 
the  hemisphere,  or  360  spher.  deg.,  by  twice  A  ABC ; 
.:  360  spher.  deg.  -{-2  A  ABC ^2(A -\- B  -{- c)  spher.  deg. ; 
.-.  AABC^{A-\-B-{-C  —  180)  spher.  deg. ; 

I.e.,  A  ^7? (7  <>^  spher.  deg.      q.e.d.     (636) 

663.  Scholium.  E,  it  must  be  remembered,  is  here  em- 
ployed as  a  numerical  measure,  an  abstract  number.  If,  for 
example,  the  angles  of  the  spherical  triangle  ABCSive  73°,  87°, 
and  100°,  respectively,  then  J^  =  73  +  87  +  100  -  180  =  80. 
Hence  we  know  that  A  ABC^SO  spherical  degrees  =o=|^To=-J^ 
of  the  surface  of  the  sphere. 


POLAR    TRIANGLES.  829 

Proposition  XXV.     Theorem. 

664.  Any  spherical  polygon  is  equivalent  to  as 
7)%any  spherical  degrees  as  there  are  angular  degrees 
in  its  spherical  excess. 


Given:  A  spherical  ^olj^on  ABODE,  whose  spherical  excess  is 
E  degrees ; 

To  Prove :  Polygon  ABODE  is  equivalent  to  E  spherical  degrees. 

Through  any  vertex  A  and  the  opposite  vertices,  describe 
arcs  of  great  circles,  AO,  AD,  dividing  the  polygon  into  spher- 
ical triangles. 

Now  the  area  of  each  triangle  is  equal  to  as  many  spheri- 
cal degrees  as  there  are  angular  degrees  in  its  spherical  excess 
(661).  Hence  the  area  of  the  polygon  is  equal  to  as  many 
spherical  degrees  as  there  are  degrees  in  the  sum  of  the 
spherical  excesses  of  the  triangles ;  that  is,  as  many  degrees 
as  there  are  in  the  spherical  excess  of  the  polygon;  (637) 
.'.  polygon  ^£C2)-£J  :o=^  spherical  degrees.       q.e.d. 

665.  Cor.  The  area  of  any  spherical  polygo7i  is  to  the  sur- 
face of  the  sphere  as  E  is  to  720. 

For  (636,  637)  E  being  the  spherical  excess  of  any  poly- 
gon P,  whether  of  three  or  more  sides,  and  S  the  surface  of 
the  sphere, 

P'.S=:E'.  720. 

Exercise  810.  If  the  angles  A,  B,  C,  D,  E,  of  the  spherical  pen- 
tagon ABODE  are  140°,  90°,  93°,  120°,  and  117°,  respectively,  what 
part  of  the  spherical  surface  is  ABODE  ? 


330  SOLID   GEOMETRY.  — BOOK  IX. 

EXERCISES. 
QUESTIONS. 

811.  What  is  the  locus  of  all  the  points  in  space  that  are  at  a  given 
distance  from  a  given  straight  line  ? 

812.  The  cylinder  may  be  regarded  as  the  limiting  form  of  what 
solid  ? 

813.  What  is  the  locus  of  all  the  straight  lines  in  space  that  make 
a  given  angle  with  a  given  straight  line  at  a  given  point  ? 

814.  The  cone  may  be  regarded  as  the  limiting  form  of  what  solid  ? 

815.  What  is  the  locus  of  all  the  points  in  space  that  are  at  a  given 
distance  from  a  given  point  ? 

816.  Two  plane  triangles  that  are  mutually  equiangular  are  not 
necessarily  equilateral ;  but  spherical  triangles  on  equal  spheres,  if 
mutually  equiangular,  are  also  mutually  equilateral.     Why  so  ? 

817.  If  straight  lines  be  drawn  from  any  point  in  a  spherical  surface 
to  the  extremities  of  a  diameter,  what  angle  will  those  lines  contain  ? 

818.  The  plane  that  is  tangent  to  a  sphere  at  a  given  point  is  the 
locus  of  what  lines  ? 

819.  What  is  the  locus  of  all  the  points  in  space  that  have  their  dis- 
tances from  two  given  parallel  lines  in  a  given  ratio  ? 

820.  What  is  the  locus  of  all  the  points  in  space  such  that  the  dis- 
tances of  each  from  a  given  straight  line  and  a  given  point  in  that  line 
have  a  given  ratio  ? 

821.  What  is  the  locus  of  all  the  points  in  space  at  a  given  distance 
from  a  given  plane  ? 

822.  What  is  the  locus  of  all  the  points  in  space  at  a  given  distance 
from  a  given  spherical  surface  ?  Under  what  circumstances  will  the 
locus  consist  of  one  surface  only  ? 

823.  What  is  the  locus  of  all  the  points  in  space  at  a  given  distance 
from  a  given  circular  cylindrical  surface  ?  Under  what  circumstances 
will  the  locus  consist  of  one  surface  only  ? 

824.  What  is  the  locus  of  all  the  points  in  space  at  a  given  distance 
from  a  given  circular  conical  surface  ? 


EXERCISES.  331 


THEOREMS. 

825.  The  locus  of  all  the  points  such  that  lines  drawn  from  it  to  the 
extremities  of  a  given  straight  line  form  a  right  angle,  is  a  spherical 
surface. 

826.  If  any  number  of  lines  in  space  pass  through  a  given  point, 
the  feet  of  the  perpendiculars  from  any  other  point  to  these  lines  lie 
upon  a  spherical  surface. 

827.  If  any  number  of  lines  in  a  plane  pass  through  a  point,  the 
feet  of  the  perpendiculars  to  those  lines  from  any  point  without  the 
plane  lie  in  a  circle. 

828.  If  from  a  point  on  the  surface  of  a  sphere  as  pole,  with  a 
polar  distance  equal  to  one  third  the  chord  of  a  quadrant,  a  circle  be 
described,  the  radius  of  this  circle  will  be  one  half  the  radius  of  the 
sphere. 

829.  Any  lune  is  to  a  trirectangular  triangle  as  its  angle  is  to  half 
a  right  angle. 

830.  Spherical  polygons  on  equal  spheres  are  as  their  spherical 
excesses. 

831.  In  any  right  spherical  triangle,  if  one  side  be  greater  than  a 
quadrant,  there  must  be  a  second  side  greater  than  a  quadrant. 

832.  In  any  right  spherical  triangle,  a  side  less  than  a  quadrant 
subtends  an  acute  angle  ;  a  side  greater  than  a  quadrant  subtends  an 
obtuse  angle. 

833.  Two  right  spherical  triangles  are  equal  or  symmetrical  if  the 
hypotenuse  and  an  adjacent  angle  of  the  one  are  severally  equal  to 
the  hypotenuse  and  an  adjacent  angle  of  the  other. 

834.  Two  right  spherical  triangles  are  equal  or  symmetrical  if  the 
hypotenuse  and  an  arm  of  the  one  are  severally  equal  to  the  hypot- 
enuse and  an  arm  of  the  other. 

835.  The  bisector  of  the  angle  contained  by  arcs  of  great  circles  is 
the  locus  of  all  points  within  the  angle  and  equidistant  from  its  sides. 


Book  X. 

MEASUREMENT  OF   THE    THREE  ROUND 
BODIES. 


^^*^< 


CYLINDERS. 

666.  A  prism  is  inscribed  in  a  cylinder  when  its  bases 
are  inscribed  in  the  bases  of  the  cylinder,  and  its  lateral 
edges  are  elements  of  the  lateral  surface  of  the  cylinder. 

667.  A  prism  is  circumscribed  about  a  cylinder  when  its 
bases  are  circumscribed  about  the  bases  of  the  cylinder. 

668.  The  lateral  area  of  a  cylinder  is  the  area  of  its 
lateral  surface. 

Proposition   I.     Theorem. 

669.  The  lateral  area  of  a  cylinder  of  revolution  is 
measured  hy  the  product  of  the  circumference  of  its 
base  hy  its  altitude. 


^.^= 

c 

A'    \      B 

^^' 

d\ 

..  c\ 

y^d 

\P 

Given :  C,  the  circumference  of  the  base  of  a  cyhnder  of  revo- 
lution ABCD-c' ;  II,  its  altitude  ;  and  s,  its  lateral  surface ; 
To  Prove :  S  is  equal  to  C  x  //. 


CYLINDERS. 


333 


Inscribe  in  the  cylinder  a  regular  prism  ABCD-C',  whose 
bases  are  regular  polygons  inscribed  in  the  bases  of  the 
cylinder. 

If  we  denote  by  s  the  lateral  surface  of  the  prisixi,  and  by 
p  the  perimeter  of  each  of  the  base  polygons;  then,  H 
being  also  its  altitude, 

s=pxH,  (514) 

whatever  be  the  number  of  lateral  faces  of  the  prism. 

Let  the  number  of  lateral  faces  be  indefinitely  increased 
by  continually  doubling  the  number  of  sides  of  the  base 
polygons. 

Then  as  p  has  for  limit  C  (392),  and  the  lateral  edges 
of  the  prism  are,  if  indefinitely  increased  in  number,  the 
elements  of  the  surface  s, 


s  has  for  limit  S  ; 
.:  S  =  C  X  H. 


Q.E.D.       (23(5) 


670.  Cor.  1.  If  the  cylinder  be  generated  by  a  rectangle 
whose  sides  are  H  and  R,  revolving  about  II,  then  H  is  the  alti- 
tude of  the  cylinder  and  R  the  radius  of  the  base.  Hence  the 
perimeter  of  the  base  is  2  7r'R  (396),  and  for  the  lateral 
area  S  we  obtain  the  expression,  S  —  27r  •  R  -  H. 

671.  CoR.  2.  Since  the  area  of  each  base  is  tt-R^  (398), 
we  obtain  for  T,  the  total  area  of  a  cylinder  of  revolution,  the 
expression,  t=2tt  -  R{H -\-  R). 


672.  Definition.  Similar  cylinders 
of  revolution  are  generated  by  similar 
rectangles  revolving  about  homolo- 
gous sides. 


673.  CoR.  3.  The  lateral  areas,  or 
the  total  areas,  of  similar  cylinders  of 
revolution,  are  as  the  squares  of  their  altitudes  or  radii 


334 


SOLID   GEOMETRY.— BOOK  X. 


For  let  S  and  S'  denote  the  lateral  areas  of  two  similar 

cylinders  of  revolution;  R  and  R\  the  radii  of  their  bases; 

H  and  h\  their  altitudes ;   T  and  T\  their  total  areas ; 

then,  since  the  generating  rectangles  are  similar,   (Hyp.) 

H:H'  =  R:R'  =  H-\-H':R  +  R';  (246) 

.-.   S:S'  =  27r-R-H:27r'R^'H^  =  H^ :  h"  =  R- :  R'% 

and    T:T'=27r'R{H+R):27r'R'{H'-{-R')  =  H':H''=R^:R'\ 

674.  Scholium.  The  lateral  area  of  any  cylinder  is  equal 
to  the  product  of  the  perimeter  of  a  right  section  of  the 
cylinder  by  an  element  of  its  surface. 

This  may  be  proved  by  a  method  similar  to  that  employed 
in  the  proof  of  Prop.  I.,  assuming  that  the  bases  of  the 
cylinders,  when  not  circular,  are  still  the  limits  of  inscribed 
polygons,  the  number  of  whose  sides  is  indefinitely  great. 


Proposition  II.     Theorem. 

675.  The  volume  of  a  cylinder  of  revolution  is  meas- 
ured hy  the  product  of  its  base  by  its  altitude. 


Given :  F,  the  volume  of  a  cylinder  of  revolution  AB,  whose 
base  is  B  and  altitude  H; 

To  Prove :  V  is  equal  lo  B  x  H. 

Let  f'  denote  the  volume  of  a  regular  inscribed  prism  of 
any  number  of  faces ;  B ',  its  base ;  H  will  also  be  its  altitude. 


CYLINDERS.  335 

Now  whatever  be  the  number  of  sides  of  the  prism, 

r  =  B'  XH.  (535) 

But  when  the  number  of  faces  is  indefinitely  increased, 
B^  has  for  limit  B,  and  F'  has  for  limit  V ; 

.'.   V  =  B  X  H.  Q.E.D.      (236) 

676.  CoR.  1.  Let  v  be  the  volume  of  the  cylinder,  R  the 
radius  of  its  base,  and  H  its  altitude ;  then,  since  B  =  ir  •  R^, 

V=7r'  R^  '  H. 

677.  CoR.  2.  The  volumes  of  similar  cylinders  of  revolution 
are  to  each  other  as  the  cubes  of  their  altitudes  or  radii. 

For  if  V  and  F'  be  the  volumes  of  two  similar  cylinders 
of  revolution,  R  and  R^  the  radii  of  their  bases,  H  and  H^ 
their  altitudes ; 

since  the  generating  rectangles  are  similar,     (Hyp.) 

H:II'=R:R'; 

.'.  V:V'  =  7r'R^'H:7r-  R''-'H=H^:H'^=R^:  R^\ 

678.  Scholium.  The  volume  of  any  cylinder  is  meas- 
ured by  the  product  of  its  base  by  its  altitude. 

This  may  be  proved  by  the  same  method  as  that  employed 
in  Prop.  II.,  making  the  assumption  before  referred  to  (674). 

Exercise  836.  Show  that,  by  the  following  construction,  two  lines, 
Jf  and  iV,  can  be  found  such  that  Jf :  iV=  355  :  113.  Take  AB=  10 
units  of  any  convenient  length,  and  on  it  lay  off  ^  C  =  5,  and  AD  =  3. 


Draw  BE  ±  to  AB  and  =  1.  Join  AE,  and  draw  CF,  DG,  each  ±  to 
AB,  and  ^meeting  AE  in  F,  G  resp.  Take  M  =  3  AB  +  AC  +  CF, 
and  N=AB-hBE+  DG.    Then  M:N=Sdd'.  113. 


336  SOLID   GEOMETRY.  — BOOK   X. 

CONES. 

679.  A  pyramid  is  inscribed  in  a  cone  when  its  base  is 
inscribed  in  tbe  base  of  the  cone,  and  its  vertex  is  that  of 
the  cone,  the  lateral  edges  of  the  pyramid  thus  being  ele-, 
ments  of  the  surface  of  the  cone. 

680.  A  pyramid  is  circicmscribed  about  a  cone  when  its 
base  is  circumscribed  about  the  base  of  the  cone,  and  its 
vertex  is  that  of  the  cone. 

681.  The  altitude  of  a  cone  is  the  perpendicular  distance 
from  its  vertex  to  its  base. 

682.  The  slant  height  of  a  cone  of  revolution  is  equal  to 
the  hypotenuse  of  the  generating  triangle. 

683.  The  lateral  area  of  a  cone  is  the  area  of  its  lateral 
surface. 


Proposition  III.     Theorem. 

684.  The  lateral  area  of  a  cone  of  revolution  is  meas- 
ured hy  the  -product  of  the  circumference  of  its  base 
hy  one  half  its  slant  height. 


Given :  S,  the  lateral  area ;    C,  the  circumference  of  the  base ; 
and  L,  the  slant  height  of  a  cone  of  revolution  S-ADF ; 
To  Prove :  S  is  equal  to  ^  C  x  L. 

Inscribe  in  the  base  any  regular  polygon  ADF;  and  upon 
this  polygon  as  base  construct  the  regular  inscribed  pyra- 


CONES.  337 

mid  S-ADF.    If  we  denote  by  s  the  lateral  area  of  the 

pyramid,  by  p  the  perimeter  of  its  base,  and  by  I  its  slant 

height, 

s^\pxl,  (549) 

whatever  be  the  number  of  lateral  faces  of  the  pyramid. 

Conceive  the  number  of  lateral  faces  to  be  indefinitely 

increased  by  continually  doubling  the  number  of  sides  of 

the  base  polygon ;    then,  since  s,  p,  and  I  have  for  limits 

S,  C,  L,  respectively, 

s  =  \c  X  L.  Q.E.D.     (536) 

685.  CoR.  1.  If  R  denote  the  radius  of  the  base,  then 
C  =  2  TT  .  /?  (396),  and  S  =  ^(2  7r  •  R  -  L)  =  7r  -  R  -  L. 

Also,  since  the  area  of  the  base  =ir  -  R^  (398),  the  total 
area,  T,  of  the  surface  of  a  cone,  is  expressed  by 

T=7r'  R-L  +  TT'  R^=7r-  R(L  +  R). 

686.  Definition.  Similar  cones  of  revolution  are  gen- 
erated by  similar  right  triangles  revolving  about  homologous 
arms  as  axes. 

687.  CoR.  2.  The  lateral  areas,  or  the  total  areas,  of  sim- 
ilar cones  of  revolution,  are  to  each  other  as  the  squares  of 
their  altitudes,  or  as  the  squares  of  their  radii. 

This  may  be  proved  as  was  Cor.  3  of  Prop.  I. 

688.  Definition.  A  truncated  cone  is  the  portion  of 
a  cone  intercepted  between  the  base  of  the  cone  and  a  plane 
cutting  its  lateral  surface. 

689.  Definition.  A  frustum  of  a  cone  is  a  truncated 
cone  that  has  the  cutting  plane  parallel  to  the  base.  The 
section  made  by  the  cutting  plane  is  the  upper  base  of  the 
frustum ;  the  perpendicular  distance  between  its  bases  is 
the  altitude  of  the  frustum;  and  the  portion  of  the  slant 
height  of  the  cone  that  is  intercepted  between  the  bases  is 
the  slant  height  of  the  frustum. 

Geom.— 22 


338  SOLID   GEOMETRY.— BOOK  X. 

Proposition  TV.     Theorem. 

690.  The  lateral  area  of  a  frustum  of  a  cone  of 
revolution  is  measured  hy  the  product  of  its  slant 
height  by  half  the  sum  of  the  circumferences  of  its 
hases. 


Given :  S,  the  lateral  surface,  C  and  c,  the  circumferences  of  its 
bases,  and  L,  the  slant  height  of  ABC-c,  a  frustum  of  a  cone  of 
revolution; 

To  Prove :  s  is  equal  io  ^L{c  -\-c). 

Inscribe  in  the  frustum  ABC-c  a  frustum  of  a  regular 
pyramid.  If  we  denote  its  lateral  surface  by  s,  the  perime- 
ters of  its  upper  and  lower  bases  by  p  and  P  respectively, 
and  its  slant  height  by  I, 

s  =  i,l{P+p),  (550) 

whatever  be  the  number  of  lateral  faces  of  the  pyramid. 

Conceive  the  number  of  lateral  faces  of  the  pyramidal 
frustum  to  be  indefinitely  increased  by  continually  doubling 
the  number  of  sides  of  its  base  polygons  ;  then. 

since  s,  p,  F,  I,  have  for  limits  S,  c,  C,  L,  respectively, 

S  =  ^L{C-{-C).  Q.E.D.       (236) 

691.  Cor.  The  lateral  area  of  a  frustum  of  a  cone  of 
revolution  is  measured  by  the  product  of  its  slant  height  by  the 
circumference  of  a  section  equidistant  from  the  bases. 


CONES.  339 

For  if  oa  be  the  radius  of  a  sec-  y, 

-^k .0 

tion   equidistant  from   the  bases, 

whose  radii  are  OA   and   O'A'  re- 
spectively, "^ 

since  oa  =  ^{OA+  O^A'),  (150) 

.-.  circumf.  oa  =  |(circumf.  OA  -f-  circumf.  O'a')  ; 
.-.  circumf.  oax^^'=^  (circumf.  0^  +  circumf.  0'a')xAA\ 

-♦ 

Proposition  V.     Theorem. 

692.  The  volume  of  a  cone  of  revolution  is  jneasured 
hy  one  third  the  product  of  its  base  by  its  altitude. 

8 


Given:  V,  the  volume,  n,  the  base,  //,  the  altitude,  of  a  cone 
of  revolution  S-ABC; 

To  Prove :  V  is  equal  to  ^  i?  x  ^. 

Inscribe  in  the  cone  a  regular  pyramid,  S-ABC;  then 
denoting  its  volume  by  v',  its  base  by  B',  H  being  also  its 
altitude, 

F'  =  i  i?'  X  //,  (555) 

whatever  be  the  number  of  lateral  faces  of  the  pyramid. 


340  SOLID   GEOMETRY.  — BOOK  X. 

Conceive  the  number  of  lateral  faces  of  the  pyramid  to  be 
indefinitely  increased ; 

since  F'  and  B^  have  for  limits  V  and  B  resp., 

V  =  \  B  X  H.  Q.E.  D.      (236> 

693.  CoR.  1.  If  R  denote  the  radius  of  the  base  ;  then,  since 
B^iT'  R\  (398) 

694.  CoR.  2.  Similar  cones  of  revolution  are  to  each  other 
as  the  cubes  of  their  altitudes,  or  as  the  cubes  of  their  radii. 

For  if  R  and  R'  are  the  radii  of  two  similar  cones  of  revo- 
lution, H  and  h'  their  altitudes,  V  and  r'  their  volumes, 

since  the  generating  triangles  are  similar,     (Hyp.) 

H:H'=R:R'; 

.-.  F:  F'  =  i  TT  .  i2-  .  iT:  i  TT  •  R"  •  H' =  H^  :  H'^  =  R^  :  R'\ 

695.  Scholium.  The  volume  of  any  cone  is  measured  by 
one  third  the  product  of  its  base  by  its  altitude. 

This  may  be  proved  by  the  same  method  as  that  em- 
ployed in  the  proof  of  Prop.  V.,  making  the  assumption 
before  referred  to  (674). 

Exercise  837.  If,  by  the  method  of  Exercise  836,  a  straight  line 
be  found  approximately  equal  to  the  circumference  of  a  circle  one 
yard  in  diameter,  by  what  fraction  of  an  inch  will  the  line  be  too 
great,  taking  only  two  significant  figures  ? 

838.  The  diameters  of  the  bases  of  a  frustum  of  a  cone  are  10  in.  and 
8  in.  respectively,  and  its  slant  height  is  12  in.     Find  its  lateral  area. 

839.  Find  the  area  of  a  section  of  that  same  cone  equidistant  from 
its  bases. 

840.  Find  the  volume  of  a  cone  of  revolution  the  radius  of  its  base 
being  10  in.  and  its  altitude  20  in. 

841.  What  is  the  altitude  of  a  similar  cone  of  twice  the  volume  ? 


CONES.  341 

Pkoposition  VI.     Theorem. 

696.  The  volume  of  a  frustum  of  a  cone  of  revolu- 
Hon  is  measured  hy  one  third  the  product  of  its  alti- 
tude hy  the  sum  of  the  hases  of  the  frustum  and  a 
mean  proportional  between  those  bases. 


Given :  r,  the  volume,  B  and  B',  the  bases,  and  H,  the  altitude, 
of  a  frustum  ABC-c; 

To  Prove :       V  is  equal  to  \  H  {B  ^  b'-  -{-  Vb  -  B') . 

Inscribe  in  the  frustum  a  frustum,  ABC-c,  of  a  regular 
pyramid ;  then,  denoting  its  volume  by  v,  its  bases  by  b  and 
b',  H  being  its  altitude, 

1?  =  1  ^  (6  +  6'  +  V5T6'),  (559) 

whatever  be  the  number  of  the  lateral  faces  of  the  frustum. 
Conceive  the  number  of  lateral  faces  of  the  inscribed 
frustum  to  be  indefinitely  increased ; 

since  v,  b,  and  b',  have  for  limits  F,  B,  and  B',  resp., 


V=ill  {B  +  B'  -\-^B-  B') .  Q.E.D.      (236) 

697.  CoR.    If  R  and  R'  denote  the  radii  of  the  bases  of  the 
frustum,  as 

B  =  TT  '  R^,  B'  =  TT  '  R'%  and  V^  •  B'  ==7r  -  R-  R', 
V=^7r'H  {R^  +  R"'-^-  R-R'). 

698.  Scholium.    The  volume  of  a  frustum  of  any  cone  is 
measured  by  one  third  the  product  of  its  altitude,  etc. 

The  same  remark  applies  here  as  in  Arts.  674  and  695. 


342  SOLID   GEOMETRY.  — BOOK  X. 


EXERCISES. 
NUMERICAL. 

842.  Find  the  lateral  area  and  the  total  area  of  a  cylinder  of  revolu- 
.tion  whose  altitude  is  18  in.  and  the  diameter  of  its  bases  12  in. 

843.  What  is  the  volume  of  the  same  cylinder  ? 

844.  The  total  area  of  a  cylinder  of  revolution  is  700  sq.  in.,  its 
altitude  is  14  in.     What  is  the  diameter  of  a  base  ? 

845.  What  should  be  the  altitude  of  a  cylinder  of  revolution,  the 
diameter  of  which  is  5  in.,  so  that  the  lateral  area  shall  be  a  square  foot  ? 

846.  What  should  be  the  diameter  of  a  cylinder  of  revolution 
whose  altitude  is  10  in.,  so  that  its  total  area  shall  be  500  sq.  in.? 

847.  The  altitude  of  a  cylinder  of  revolution  is  three  times  its  diame- 
ter ;  the  total  area  is  1200  sq.  in.     Find  the  altitude  and  diameter. 

848.  The  altitudes  of  two  similar  cylinders  of  revolution  are  as 
7  to  5.     What  is  the  ratio  of  their  total  areas  ?     Of  their  volumes  ? 

849.  What  formula  expresses  the  total  area  of  a  cylinder  of  revolu- 
tion whose  altitude  and  radius  are  equal  ? 

850.  What  formula  expresses  the  volume  of  the  same  cylinder  ? 

851.  What  is  the  ratio  of  the  volume  of  the  same  cylinder  to  the 
volume  of  a  cube  having  the  same  altitude  ? 

852.  Find  the  lateral  area  and  the  total  area  of  a  cone  of  revolu- 
tion whose  altitude  is  15  in.,  and  the  diameter  of  whose  base  is  12  in. 

853.  Find  the  volume  of  the  same  cone. 

854.  The  total  area  of  a  cone  of  revolution  is  400  sq.  in. ;  its  alti- 
tude is  10  in.     What  is  the  diameter  of  its  base  ? 

855.  What  should  be  the  altitude  of  a  cone  of  revolution  whose 
base  has  a  diameter  of  10  in.,  so  that  the  lateral  area  may  be  a  square 
foot? 

856.  What  should  .be  the  radius  of  the  base  of  a  cone  of  revolution 
whose  altitude  is  10  in.,  so  that  its  total  area  shall  be  100  sq.  in.  ? 

857.  The  altitude  of  a  cone  of  revolution  is  four  times  the  radius  of 
its  base ;  the  lateral  area  is  500  sq.  in.  Find  the  radius  and  alti- 
tude. 


EXERCISES.  343 

858.  The  altitudes  of  two  similar  cones  of  revolution  are  as  11  to  8. 
Wliat  is  the  ratio  of  tlieir  total  areas  ?     Of  their  volumes  ? 

859.  What  formula  expresses  the  total  area  of  a  cone  of  revolution 
whose  altitude  is  equal  to  tlie  radius  of  its  base  ? 

860.  What  formula  expresses  the  volume  of  the  same  cone  ? 

861.  What  is  the  ratio  of  the  volume  of  the  same  cone  to  the  vol- 
ume of  a  regular  tetrahedron  liaving  the  same  altitude  ? 

862.  What  should  be  tlie  altitude  of  such  a  cone,  that  its  lateral 
area  may  be  100  sq.  in.  ? 

863.  What  should  be  the  altitude  of  such  a  cone,  that  its  volume 
may  be  1000  cu.  in.  ? 

864.  What  is  the  lateral  area  and  the  total  area  of  a  frustum  of  a 
cone  of  revolution  whose  altitude  is  20  in.,  and  the  diameters  of  whose 
bases  are  6  in.  and  14  in,  respectively  ? 

865.  What  is  the  volume  of  the  same  frustum  ? 

866.  The  diameters  of  the  bases  of  a  frustum  of  a  cone  of  revolution 
are  10  in.  and  16  in.  respectively  ;  its  volume  is  575  cu.  in.  What  is 
its  altitude  ? 

867.  How  far  from  the  base  must  a  cone,  whose  altitude  is  16  in., 
be  cut  by  a  plane  so  that  the  frustum  shall  be  equivalent  to  one  half 
the  cone  ? 

868.  What  is  the  ratio  of  the  lateral  surfaces  of  a  right  circular  cyl- 
inder and  a  right  circular  cone  of  the  same  base  and  altitude,  if  the 
altitude  is  three  times  the  radius  of  the  base  ? 

869.  The  diameter  of  a  right  circular  cylinder  is  10  ft.,  and  its  alti- 
tude 7  ft.    What  is  the  side  of  an  equivalent  cube  ? 

870.  The  altitude  of  a  cone  of  revolution  is  15  in.,  and  the  radius 
of  its  base  5  in.  What  should  be  the  diameter  of  a  cylinder  of  revo- 
lution liaving  the  same  altitude  and  lateral  area  ? 

871.  What  should  be  tlie  ratio  of  the  exterior  to  the  interior  diame- 
ter of  a  hollow  cylinder  of  revolution,  so  that  it  shall  contain  one  half 
the  volume  of  a  solid  cylinder  of  the  same  dimensions  ? 

872.  In  order  that  a  cylindrical  tank  with  a  depth  of  12  ft.  may 
contain  2000  gal.,  what  should  be  its  diameter  ? 

873.  How  many  cubic  inches  of  iron  would  be  required  to  make 
that  tank,  its  walls  being  one  third  of  an  inch  thick  ? 


344 


SOLID   GEOMETRY.  — BOOK  X, 


SPHERES. 
Proposition  VII.     Theorem. 

699.  The  area  of  the  surface  generated  by  a  straight 
line  revolving  ohout  an  axis  in  its  -plane,  is  measured 
by  the  product  of  the  projection  of  that  line  upon  the 
axis  by  the  circumference  of  the  circle  whose  radius 
is  the  perpendicular  froin  the  axis  to  the  mid  point 
of  the  line. 


DO 


Given:   ah,  the  projection  upon  XF  of  AB  revolving   about 
XY,  and  OP  ±  to  ^i?  at  its  mid  point,  and  meeting  xrin  0; 
To  Prove :     Area  generated  by  ^5  is  equal  io  ah  x2  it  -  OP. 


Draw  PD  1.  to  XY,  and  AC  W  to  XY.  - 
Since  the  surface  generated  hj  AB  is  the  lateral  surface 

(576) 

(689) 
(292) 
(285) 
(237) 


of  a  frustum  of  a  cone, 

area  gen.  hj  AB  =  ab  x  2  tt  •  PD. 
Now  A  ABC  is  similar  to  A  POD; 

.:  AB:  OP  =  AC:  PD; 
.'.  AB  X  PD  =  AC  X  OP  =  ah  X  OP; 


.:  AB  X  2  TT  •  PD  =  ah  X  2  77  •  OP; 
i.e.,  area  generated  hy  AB  =  ah  x  2  tt  -  OP.  q.e.d. 

If  ^j5  meets  XY,  the  surface  generated  is  still  a  conical 
surface  whose  area=a6  x  2  tt  •  OP,  as  follows  from  Prop.  III. 

If  ^P  is  parallel  to  XY,  the  surface  generated  is  a  cylin- 
drical surface  whose  area  =  a6  x  2  tt  •  OP,  as  follows  from 
Prop.  I. 


SPHERES. 


345 


Proposition  VIII.     Theorem. 

700.  The  area  of  the  surface  of  a  sphere  is  meas- 
ured hy  the  product  of  its  diameter  hy  the  circum- 
ference of  a  great  circle. 


(699) 


Given :  S,  the  surface  of  a  sphere  generated  by  the  revolution 
of  the  semicircle  ABODE  about  the  diameter  AOU,  where  OA 
equals  R; 

To  Prove :    Area  of  s  is  equal  to  AR  x  2  tt  -  R. 

Inscribe  in  the  semicircle  a  regular  semipolygon  AB ...  E, 
of  any  number  of  sides,  and  draw  Bb,  Co,  Bd,  Js  to  ^^. 

From  0  draw  OP  _L  to  AB.  Then  OP  bisects  AB  (172), 
and  is  equal  to  each  of  the  Js  drawn  from  0  to  the  equal 
chords  BC,  CD,  DE  (182). 

Now  area  AB  =  Ab.  x  2  tt  •  OP, 
area  BC  =  bc  x  2  tt  -  OP, 
area  CD  =  cd  x  2  tt  -  OP,  etc. ; 
.-.  if  S'  denote  the  surface  generated  by  the  semipolygon, 
S'  =  {Ab  -{-bc  +  cd-\-  dE)  x2Tr'OP  =  AEx2ir'OP. 
Conceive  the  number  of  sides  of  the  semipolygon  to  be 
indefinitely  increased.      Then,  as   OP  has  for  limit  R,  the 
semipolygon  for  limit  the  semicircle,  and  -S'  for  limit  S, 

S=  AE  X2  TT'  R.  Q.E.D.      (236) 

701.  Cor.  1.  Tlie  surface  of  a  sphere  is  equivalent  to  four 
great  circles. 

'Fov  8=2  Rx2tt'  R=4.Tv  R\  (700) 

and  TT  •  ii-  is  the  area  of  a  great  circle.  (398) 


346  SOLID   GEOMETRY.  — BOOK  X. 

702.  Cor.  2.  The  areas  of  the  surfaces  of  two  spheres  are 
to  each  other  as  the  squares  of  their  radii  or  diameters. 

703.  Definitions.  A  zone  is  a  portion  of  the  surface  of  a 
sphere  included  between  two  parallel  planes.  The  altitude 
of  the  zone  is  the  perpendicular  distance  between  the  paral- 
lel planes.  The  bases  of  the  zone  are  the  circumferences  of 
the  bounding  circles.  The  zone  is  called  a  zone  of  one  base, 
if  one  of  the  parallel  planes  is  tangent  to  the  sphere ;  that 
is,  a  zone  of  one  base  is  the  surface  cut  off  by  a  plane. 

704.  Cor.  1.  The  area  of  a  zone  is  measured  by  the 
product  of  its  altitude  by  the  circumference  of  a  great  circle. 

For  (see  diagram  for  Prop.  VIII.)  the  area  of  the  zone 
generated  by  the  revolution  of  the  arc  5(7.=  6c  x  2  tt  •  i?. 

705.  Cor.  2.  Zones  on  the  same  sphere,  or  on  equal 
spheres,  are  to  each  other  as  their  altitudes. 

706.  Cor.  3.  The  area  of  a  zone  of  one  base  is  meas- 
ured by  the  area  of  the  circle  whose  radius  is  the  chord  of 
the  generating  arc. 

For  the  arc  AB  generates  a  zone  of  one  base  whose  area  is 

Ab  X2  IT  '  R  =  Tr  '  Abx  AE^TT'AB',  (398) 

{^VHQQ  AB  X  AE  =  AB  .) 

Exercise  874.  The  diameter  of  a  sphere  being  1  ft.,  what  is  the 
area  of  a  great  circle  of  that  sphere,  and  of  the  sphere  itself  ? 

875.  If  the  area  of  its  surface  is  400  sq.  in.,  what  is  the  diameter 
of  the  sphere  ? 

876.  What  is  the  ratio  of  the  surfaces  of  two  spheres  whose  radii 
are  10  in.  and  12  in.  respectively  ? 

877.  What  is  the  area  of  a  zone  of  a  sphere  12  in.  in  diameter,  the 
altitude  of  the  zone  being  3  in.  ? 

878.  What  fraction  of  the  diameter  of  a  sphere  should  the  altitude 
of  a  zone  be  so  as  to  contain  -  th  of  the  surface  ? 


SPHERES.  347 

Proposition  TX.     Theorem. 

707.  The  volume  of  a  sphere  is  measured  hy  one 
third  the  product  of  its  surface  hy  its  radius. 


Given :  V,  the  volume,  and  R,  the  radius,  of  the  sphere  whose 
surface  is  S; 

To  Prove :  V  is  equal  to  ^  .s  x  i?. 

Conceive  a  great  number  of  points  to  be  taken  on  the 
surface  of  the  sphere,  and  join  them,  two  and  two,  by  arcs 
of  great  circles,  so  as  to  form  on  the  surface  a  network  of 
triangles  such  as  ABC. 

The  planes  of  these  arcs  will,  by  their  intersections,  form 
the  lateral  faces  of  triangular  pyramids,  such  as  0-ABC, 
having  their  vertices  at  0,  and  their  bases  plane  triangles 
whose  vertices  coincide  with  those  of  the  spherical  triangles. 

Since  the  volume  of  each  pyramid  is  equal  to  one  third 
the  product  of  its  base  and  altitude  (555),  if  we  denote  by 
V'  the  sum  of  these  volumes,  by  b,  b',  b",  •••  the  bases,  and 
by  h,  h',  h",  •••  the  altitudes,  we  find  , 

V'  =  ^{b'h-\-b'  'h'  +  b"  .  h"  -{-  ...). 

Conceive  the  number  of  triangles  to  be  indefinitely 
increased ;  then,  as  V'  has  for  limit  V,  the  sum  of  the  bases 
has  for  limit  S,  and  each  altitude  has  for  limit  R, 

V=}^S'R.  Q.E.D.       (236) 


348  SOLID   GEOMETRY.  —  BOOK  X. 

708.  Cor.  1.   Since   s==4:7r'R^,  (700) 

F=  i  X  47r  .  i^^  •  i?  =  Itt  •  7?^  =  i-TT  •  Z)l 

709.  Cor.  2.  The  volumes  of  spheres  are  to  each  other  as 
the  cubes  of  their  radii  or  diameters. 

For  since  v^^tt-  R^  =  ^ttD^  and  v' =  ^ttR'^  =  ^tt  -  D'% 

710.  Definitions.     A  spherical  sector  is  a  portion  of  a 

sphere  generated  by  a  sector  of  the 
semicircle  that  generates  the  sphere. 
Thus  the  revolution  of  the  sector 
AOB  generates  the  spherical  sector 
0-ABB'.  The  revolution  of  the 
sector  OCD  generates  the  spherical 
sector  0-GDD^C\  The  base  of  the 
spherical  sector  is  the  zone  gener- 
ated by  the  circular  sector. 

711.  Cor.  1.  The  volume  of  a 
spherical  sector  is  measured  by  one  third  the  product  of  the 
zone  that  forms  its  base,  by  the  radius  of  the  sphere. 

In  the  case  of  the  spherical  sector  0-A  B  B',  it  is  manifest, 
that  whatever  part  the  base  bb'  is  of  the  surface  of  the 
sphere,  that  same  part  is  0-ABB'  of  the  sphere.  In  the 
case  of  0-CBD'g',  we  see  that  this  sector  consists  of  the 
difference  of  the  spherical  sectors  0-CA'c'  and  0-DA'b',  and 
its  base  consists  of  the  difference  of  the  portions  of  the 
surface  of  the  sphere  that  are  cut  off  by  C&  and  DZ)'  respec- 
tively. If  V  denote  the  volume  of  the  sector,  z  the  area  of 
the  zone,  and  H  its  "altitude,  then 

712.  CoR.  2.  The  volumes  of  spherical  sectors  on  equal 
spheres  are  as  the  zones  that  form  their  bases,  or  as  their 
altitudes. 


SPHERES.  349 

For  since    V=^Z' R^^tt- R^-H,  Smd  V'  =  l  Z' •  R' =z^7r' r'^-h, 
V:  r'  =  Z:  Z'  =  H:  H'. 

713.  Definition.  A  spherical  pyramid  is  a  solid  bounded 
by  a  spherical  polygon  as  base,  whose  vertex  is  the  center  of 
the  sphere,  and  whose  sides  are  the  sides  of  the  polyhedral 
angle  formed  by  the  planes  of  the  sides  of  the  polygon. 

714.  CoR.  1.  The  volume  of  a  spherical  pyramid  is  meas- 
ured by  one  third  the  product  of  its  base  by  the  spherical  radius. 

For  it  is  evident  that,  whatever  portion  the  base  of  the 
pyramid  is  of  the  surface  of  the  sphere,  the  same  portion 
will  the  volume  of  the  pyramid  be  of  the  volume  of  the 
sphere. 

715.  CoR.  2.  Spherical  pyramids  on  equal  spheres  are  to 
each  other  as  their  bases. 

716.  Definitions.  A  spherical  segment  is  a  portion  of  a 
sphere  included  between  two  parallel  planes.  The  altitude 
of  the  segment  is  the  perpendicular  distance  between  the 
parallel  planes ;  the  bases  are  the  sections  of  the  sphere 
made  by  those  planes.  If  one  of  the  bounding  planes  is 
tangent  to  the  sphere,  the  segment  is  said  to  be  a  segment 
of  one  base. 

Exercise  879.    What  is  the  volume  of  a  sphere  one  foot  in  diameter  ? 

880.  Find  the  diameter  of  a  sphere  whose  volume  is  one  cubic  foot. 

881.  Find  the  surface  of  a  sphere  whose  volume  is  one  cubic  foot. 

882.  The  diameters  of  two  spheres  are  10  in.  and  12  in.  respectively. 
Find  the  ratio  (1)  of  their  surfaces  ;  (2)  of  their  volumes. 

883.  One  sphere  has  twice  the  volume  of  another.  Find  the  ratio  of 
the  radius  of  the  first  to  the  radius  of  the  second. 

884.  The  altitude  of  the  base  of  a  spherical  sector  of  a  sphere  10  in. 
in  diameter  is  2  in. ;  find  the  volume  of  the  sector. 

885.  A  spherical  pyramid  has  for  base  a  trirectangular  triangle. 
What  fraction  is  the  pyramid  of  the  sphere  ? 


350  SOLID    GEOMETRY.  — BOOK  X. 

Proposition  X.     Theorem. 

717.  The  volume  of  a  spherical  segment  is  meas- 
ured hy  one  half  the  product  of  the  sum  of  its  bases 
by  its  altitude,  plus  the  volume  of  a  sphere  of  luhich 
that  altitude  is  the  diameter* 


Given:  V,  the  volume  of  the  segment  generated  by  the  revo- 
lution about  il/OJV  of  ABCD,  where  AD  =  r,  Ba  =  r',  and 
CD=  OC—  OD  =  h; 

To  Prove :  V=^h {irr^  +  tt?-'-)  +  i ttIA 

Join  OA  and  OB.  The  solid  generated  hj  ABCD  consists 
of  the  spherical  sector  generated  by  GAB,  plus  the  cone 
generated  by  OBC,  minus  the  cone  generated  by  GAB; 

.'.    V=^7r7i  ■  R'  +  ^TT  •  BC' '  GO  —  ^TT  '  Iff  •  GD      (711 ;   692) 

=  i7rj2  7?--/i  +  (7?'—  dc^)GC-{R'—  Gb')GD\  (347) 

=  \Tr\2R^'-h  -\-R\GC  -  GD)  -  (GC^  -  off)  \ 

=  1  TT^fS  R'  —  {0C~  4-  GC  •  GD  -j-  01?)\. 

But  since  {GC  -GD)-  =  h\  (Hyp.) 

OC'  ~\-  GC-  GD-{-  G~If  =  |(  OC^  4-  cff)  —  — 


=  \  ll{Tzr  -f-  7rr'-)  -|-  \  irh^.  Q.E.D. 


SPHERES. 


351 


718.  Cor.  The  volume  of  a  spherical  segment  of  one  base 
is  measured  by  one  half  the  volume  of  the  cylinder  haviyig  the 
same  base  and  altitude,  plus  the  volume  of  the  sphere  having 
that  altitude  for  diameter. 

For  when  r'  =  0,  i.e.,  when  the  segment  has  but  one  base, 
then 

F=i7rr7i+i7r/l'l 


Proposition  XI.     Theorem. 

719.  T7^e  total  surface  and  the  volume  of  a  cylinder 
circumscribed  about  a  sphere,  are  respectively  to  the 
surface  and  volume  of  the  sphere  as  3  is  to  2. 


Given:  A  cylinder  AD^,  circumscribed  about  a  sphere  bece' ; 
To  Prove :  Surf.  AI)' :  surf.  BECE'=yo\.  AI)':  vol.  BECE' =3 : 2. 

For  we  may  suppose  the  sphere  and  the  cylinder  to  be 
generated  respectively  by  the  revolution  about  J5C  of  the 
semicircle  BEC  and  the  rectangle  BD  that  is  circumscribed 
about  BEC.     Then 

surf.  AI)'  =  27r'  AB(AB  -\- BC)  =  ^tt  •  AB^; 


surf.  BECE'  =  4  TT  •  ab' 


vol.  AD' 


tT'AB    'BC 


=  4.7r'AB' 

=  2  TT  •  AB' 


YOl.BECE'  =  \Tr'BC 


AB  . 


(669) 
(700) 
(674) 
(708) 


352  SOLID   GEOMETRY,  — BOOK  X. 

Hence,  making  the  necessary  simplifications,  we  have 

surf.  AD^ :  surf.  BECE'  =  vol.  AD' :  vol.  BECE'  =  3:2.     q.e.d. 

This  interesting  theorem  is  known  as  the  Theorem  of 
Archimedes,  it  having  been  discovered  by  that  celebrated 
geometer. 

720.  Scholium.  If  we  have  a  cone  having  the  same  base 
and  altitude  as  the  cylinder  circumscribed  about  the  sphere, 
then,  since  the  volume  of  such  a  cone  is  ^ir  -  AB  •  BC  = 
J  IT  •  AB  ,  we  obtain  the  relation : 

cylinder :  sphere  :  cone  =  3:2:1. 


EXERCISES. 
NUMERICAL. 

886.  Find  the  area  of  the  surface  of  a  sphere  whose  radius  is  3|  in. 

887.  The  surface  of  a  sphere  is  to  be  100  sq.  in.  What  radius 
should  be  taken  ? 

888.  Two  spheres  have  radii  of  9  in.  and  5  in.  respectively.  What 
is  the  ratio  of  the  surfaces  of  those  spheres  ?    Of  their  volumes  ? 

889.  The  areas  of  the  surfaces  of  two  spheres  are  as  125  to  27. 
What  is  the  ratio  of  their  diameters  ?     Of  their  volumes  ? 

890.  Two  parallel  planes  intersect  a  sphere  of  18  in.  radius  at  dis- 
tances of  9  in.  and  13  in.,  respectively,  from  the  center.  Find  the 
area  of  the  intercepted  zone. 

891.  What  is  the  volume  of  the  spherical  sector  that  has  for  base 
the  zone  just  mentioned  ? 

892.  Find  the  altitude  of  a  zone  whose  area  is  100  sq.  in,  on  the 
surface  of  a  sphere  of  12  in.  radius. 

893.  In  the  same  sphere,  what  is  the  altitude  of  a  zone  that  con- 
tains one  fourth  of  the  surface  of  the  sphere  ? 

894.  What  is  the  volume,  of  a  sphere  whose  diameter  is  (1)  1  ft.  ; 
(2)  18  in.  ? 


EXERCISES.  353 

895.  The  surface  of  a  sphere  is  64  sq.  in.     Find  its  volume. 

896.  The  volume  of  a  sphere  is  5  cu.  ft.  Find  its  diameter  and 
surface. 

897.  Find  the  difference  of  the  volumes  of  two  spheres  whose 
radii  are  12  in.  and  7  in.  respectively. 

898.  The  volumes  of  two  spheres  are  as  27  to  8.  Find  the  ratio 
(1)  of  their  diameter ;  (2)  of  their  surfaces. 

899.  The  radii  of  two  spheres  are  as  4  to  5.  Find  the  ratio  (1)  of 
their  volumes  ;  (2)  of  their  surfaces. 

900.  In  a  sphere  whose  radius  is  6  in.,  find  the  altitude  of  a  zone 
whose  area  shall  be  that  of  a  great  circle. 

901.  The  area  of  a  zone  forming  the  base  of  a  spherical  sector  is 
60  sq.  in.  ;  the  radius  of  the  sphere  is  12  in.  Find  the  altitude  of  the 
zone  and  the  volume  of  the  sector, 

902.  The  volume  of  a  spherical  sector  is  25  cu.  in.  ;  the  diameter  of 
the  sphere  is  14  in.  Find  the  area  of  the  zone  that  forms  the  base  of 
the  sector. 

903.  The  altitude  of  a  cylinder  circumscribing  a  sphere  is  5  in. 
Find  the  surface  and  volume  of  the  sphere, 

904.  The  volume  of  a  sphere  is  one  cubic  foot.  Find  the  surface  of 
the  circumscribing  cylinder. 

905.  The  surface  and  volume  of  a  sphere  are  expressed  by  the  same 
number.     Find  its  diameter. 

906.  Find  the  volume  of  a  sphere  inscribed  in  a  cube  whose  volume 
is  1331  cu.  in. 

907.  Find  the  surface  of  a  cube  circumscribed  about  a  sphere 
whose  surface  is  150  sq.  in, 

.    908.   If  a  spherical  shell  have  an  exterior  diameter  of  12  in.,  what 
should  be  the  thickness  of  its  wall  so  that  it  may  contain  696.9  cu,  in.? 

909.  If  an  iron  sphere,  6  in.  in  diameter,  weigh  n  lbs.,  what  will 
be  the  weight  of  an  iron  sphere  whose  diameter  is  8  in.  ? 

910.  In  a  sphere  10  in.  in  diameter,  the  radius  of  the  lower  base 
of  a  spherical  segment  is  8  in.  Find  the  volume  of  the  segment,  its 
altitude  being  2  in.' 

Geom,  —  23 


354  SOLID   GEOMETRY.  — BOOK  X. 


THEOREMS. 

911.  The  lateral  area  of  a  cylinder  of  revolution  is  equal  to  the 
area  of  a  circle  whose  radius  is  a  mean  proportional  between  the 
altitude  and  diameter  of  the  cylinder. 

912.  The  lateral  areas  of  the  two  cylinders  generated  by  revolving 
a  rectangle  successively  about  each  of  its  containing  sides,  are  equal. 

913.  If  the  containing  sides  of  the  above  rectangle  are  as  m  is  to  n, 
the  total  areas,  and  also  the  volumes,  of  the  cylinders  generated,  will 
be  as  n  is  to  m. 

914.  If  the  slant  height  of  a  cone  of  revolution  is  equal  to  the 
diameter  of  its  base,  its  total  area  is  to  that  of  the  inscribed  sphere 
as  9  is  to  4. 

915.  The  arms  of  a  right  triangle  are  a  and  6.  Find  the  area  of  the 
surface  generated  by  revolving  the  triangle  about  its  hypotenuse. 

916.  An  equilateral  triangle  revolves  about  one  of  its  altitudes. 
What  is  the  ratio  of  the  lateral  surface  of  the  generated  cone  to  that 
of  the  sphere  generated  by  the  circle  inscribed  in  the  triangle  ? 

917.  An  equilateral  triangle  revolves  about  one  of  its  altitudes. 
Compare  the  volumes  generated  by  the  triangle,  the  inscribed  circle, 
and  the  circumscribed  circle  respectively. 

918.  A  circle  of  cardboard  being  given,  what  is  the  angle  of  the 
sector  that  must  be  cut  from  it  so  that  with  the  remainder,  a  cone  with 
a  vertical  angle  of  90°  may  be  formed  ? 

919.  If  the  diameter  of  a  sphere  be  divided  by  a  perpendicular 
plane  in  the  ratio  m  to  /^,  the  zones  thus  formed  will  also  be  as  m  to  n. 

920.  The  volume  of  a  cylinder  of  revolution  is  equal  to  one  half 
the  product  of  its  lateral  area  by  the  radius  of  its  base. 

921.  If  the  altitude  of  a  cylinder  of  revolution  is  equal  to  the 
diameter  of  its  base,  its  volume  is  equal  to  one  third  the  product  of 
its  total  area  by  the  radius  of  its  base. 

922.  The  base  of  a  cone  is  equal  to  a  great  circle  of  a  sphere,  and 
the  altitude  is  equal  to  a  diameter  of  the  sphere.  What  is  the  ratio  of 
their  volumes  ? 

923.  The  volume  of  a  sphere  is  to  that  of  the  inscribed  cube  as; 
TT  is  to  2  -=-  Vs. 

924.  A  sphere  is  to  the  circumscribed  cube  as  tt  is  to  6. 


APPENDIX 


>>*ic 


SYMMETRY. 


p' 


I.   SYMMETRY  WITH  RESPECT  TO  A  CENTER. 

721.  Definition.  Two  points  are  said  to  be  symmetrical 
with  resj^ect  to  a  third  point,  if  this  point  bisects  the  line 
joining  the  two  points.  p. 

Thus  the  points  P  and  P'  are   ^~ 
symmetrical  with  respect  to  O,  if  the  line  PP^  is  bisected 
in  O.     The  point  0  is  then  called  the  center  of  symmetry. 

722.  Definition.    Two  figures  are  said  to  be  symmetrical 
with  respect  to  a  point,  called  their 
center  of  symmetry,  if  every  point 
in   the  one  has   its    symmetrical 
point  in  the  other. 

Thus  the  figures  ABC,  A'b'c',  are 
symmetrical  with  respect  to  the 
center  0,  if  every  point  in  ABC  has 
its  symmetrical  point  in  A'b'c'. 

723.  Definition.  In  symmetrical  figures,  sides  whose 
extremities  are  mutually  symmetrical  are  said  to  be  homolo- 
gous. Thus  ^  C  is  homologous  to  A'c',  since  A  is  symmetrical 
with  A',  and  C  with  c'. 


Exercise  925.   The  opposite  vertices  of  a  regular  polygon  of  an 
even  number  of  sides  have  a  common  center  of  symmetry. 

926.  The  opposite  vertices  of  a  parallelopiped  have  a  common  cen- 
ter of  symmetry. 


365 


356  GEOMETRY.  — APPENDIX, 

Pkoposition  I.     Theorem. 

724.  If  tivo  polygons  are  syinjnetrical  tuith  respect 
to  a  center,  any  two  homologoivs  sides  are  equal  and 
parallel,  and  drawn  in  opposite  directions. 


:.0..- 


Given:  Two  polygons  AB"-E,  A'b' "  e',  symmetrical  with 
respect  to  o  ; 

To  Prove:  AB,  BC,  etc.,  are  resp.  =  and  II  to  A'b',  b'c',  etc. 

Since  OA  =  OA',  OB  =  OB',  (Hyp.) 

and  Zaob  =  Z  a'ob',  (50) 

Aaob  =  Aa'ob';  (66) 

.*.  AB  =  A'B',  and  Z  GAB  =  Z  oa'b' ;  (70) 

.-.  AB  is  II  to  A'B'-;  (110) 

also  AB,  a'b',  are  drawn  in  opposite  directions ;  (115) 

and  similarly  for  BC  and  b'c',  CD  and  c'd',  etc.  q.e.d. 

725.  Cor.  1.  Any  line  mm',  intercepted  between  two  homolo- 
gous sides,  AE,  A'e',  and  passing  through  0,  is  bisected  in  0. 

For  since  AE  is  II  to  A'e',  the  triangles  AOM,  A'om',  are 
equiangular;  and  OA=OA' ;  .-.  0M=  OM'  (70). 

726.  Cor.  2.  If  two  polygons  have  their  sides  respectively 
equal  and  parallel,  and  drawn  in  opposite  directions,  they 
have  a  center  of  symmetry. 

For  \i  AB  and  A'B'  are  equal  and  parallel,  and  drawn  on 
opposite  sides  of  A  A',  BB',  then  A  A',  BB',  are  diagonals  of 


SYMMETRY. 


357 


what  could  be  made  a  parallelogram  (142)  ;  hence  A  A',  bb', 
bisect  each  other  in  0  (146). 

-  727.  Scholium.  When  two  polygons  are  symmetrical 
with  respect  to  a  center,  one  can  be  made  to  coincide  with 
the  other  by  revolving  it  about  the  center  through  two 
right  angles  in  their  common  plane. 

728.  Definition.     A  figure  is  symmetrical  with  respect 
to  a  point,   if  every  intercept 

that  passes  through  the  point 
is  bisected  there. 

Thus  the  figure  AB  '••  cd  is 
symmetrical  with  regard  to  6, 
if  every  intercept,  as  MM',  that 
passes  through  0,  is  bisected  in  0. 

II.    SYMMETRY  WITH  RESPECT  TO  AN  AXIS. 

729.  Definition.    Two  points  are  said  to  be  symmetrical 
with  respect  to  a  straight  line,  if 

this  line  bisects  at  right  angles 
the  straight  line  joining  the  two 
points. 

Thus  the  points  P  and  P'  are 
symmetrical  with  respect  to  X  Y, 

if  XF  bisects  PP' at  right  angles.  The  line  xr  is  then 
called  the  aons  of  symmetry  as  regards  P  and  P'. 


Exercise  927.    A  circle  is  symmetrical  with  respect  to  what  point  ? 

928.  A  parallelogram  is  symmetrical  with  respect  to  what  point  ? 

929.  A  trapezium  has  no  center  of  symmetry. 

930.  Every  regular  polygon  of  an  even  number  of  sides  has  a  cen- 
ter of  symmetry. 

931.  The  axis  of  symmetry  of  the  extremities  of  a  chord  is  what 
line? 

932.  The  axis  of  symmetry  of  opposite  vertices  of  a  square  is  what 
line  ? 


358 


GEO  ME  TR  Y.  —  APPENDIX. 


A B 


730.  Definition.    Any  two  figures  are  said  to  be  symmet- 
rical  with    respect  to  an  axis,  if 
every   point   in    the    one    has   a 
point  in  the   other   symmetrical 
with  respect  to  that  axis. 

Thus  the  figures   ABC,  A'b'c', 
are  symmetrical  with  respect  to  ^ 

XY,  if  corresponding  to  every  point  in  ^5 C  there  is  a  point 
in  A'b'c'  symmetrical  with  respect  to  X  Y. 

731.  Scholium.  It  is  obvious  that,  if  the  portion  of  the 
plane  above  XT  be  revolved  about  XF  as  an  axis,  till  it  coin- 
cides with  the  portion  of  the  plane  below  X  Y,  the  figure  ABC 
will  coincide  with  a'b'c',  since  the  homologous  points  are  at 
equal  distances  from  X  F. 

732.  Definition.  A  plane  fig- 
ure is  symmetrical  loith  respect  to 
an  axis,  if  the  axis  divides  the 
figure  into  two  symmetrical  fig- 
ures. Thus  the  figure  AB ---  B' 
is  symmetrical  with  regard  to 
X  F  if  its  homologous  points  are  symmetrical  with  respect 
to  XY. 


Exercise   933.    How  many  axes  of  symmetry  may  a  circle  have  ? 
What  common  axis  of  symmetry  have  two  circles  ? 

934.  An  isosceles  triangle  is  symmetrical  with  respect  to   which 
altitude  ? 

935.  An  equilateral  triangle  has  how  many  axes  of  symmetry  ? 

936.  How  many  axes  of  symmetry  may 
be  drawn  for  (1)  a  square  ?  (2)  A  rhombus  ? 
(3)  A  regular  pentagon?  (4)  A  regular  hexa- 
gon ?  (5)  A  regular  polygon  of  2  w  sides  ? 
(6)  Of  2  w  +  1  sides  ? 

937.  In  a  quadrilateral  ABCD,  AB=AD, 
and  GB  =  CD.  Show  that  AC  is  an  axis  of 
symmetry,  and  is  perpendicular  to  BD. 


SYMMETRY. 


359 


Proposition  II.     Theorem. 

733.  If  a  figure  is  symmetrical  with  respect  to  tivo 
axes  at  right  angles  to  each  other,  it  is  also  syimnet- 
rical  with  respect  to  their  intersection  as  center. 


F 

1 

E 

G 

,J^. 

Q \^ 

'■■■■..  ;s 

D 

M 

O 

'>y 

C 

A 

7^ 

Given:  AB  •"  II,  symmetrical  with  respect  to  axes  XX',  YY', 
intersecting  in  0; 

To  Prove :  AB  •-•  iiis  symmetrical  with  respect  to  0  as  center. 
From  P,  any  point  in  the  perimeter  of  the  figure, 
draw  PQIi  ±  to  YY',  and  through  E  draw  RSP'  ±  to  XX'. 
Join  OP,  OP',  and  QS. 

Since  p  Q  =  Q  r,  (Hyp.) 

SiiidOS=QR,  (13G) 

PQ=OS.  (Ax.  1) 

Also  PQ  is  II  to  OS;  •       (106) 

.-.   OP  is  II  and  =  to  QS.  (13G) 

In  the  same  way  it  may  be  proved  that 
OP'  is  II  and  =  to  QS; 
.'.  pop'  is  3i  straight  line,  and  is  bisected  in  o; 
i.e.,  0  is  the  center  of  symmetry  of  AB  •"  ii.    q.e.d. 

734.  Scholium.  The  axes  xx',  YY',  evidently  divide  the 
figure  into  four  equal  parts.  Any  one  of  these  parts  may 
be  made  to  coincide  with  either  of  the  adjacent  parts  by 
revolving  it  about  one  of  the  axes,  or  may  be  made  to  coin- 
cide with  the  opposite  part  by  revolving  it,  in  the  plane  of 
the  figure,  through  two  right  angles. 


360 


GEOMETRY.—  APPENDIX. 


SYMMETRICAL  POLYHEDRONS. 

I.   SYMMETRY  WITH   RESPECT  TO  A  CENTER. 

735.  Definition.  Two  polyhedrons  are  said  to  be  symmet- 
rical with  respect  to  a  center,  when  each  vertex  of  the  one 
has  its  symmetrical  vertex  on  the  other  polyhedron. 


Thus  in  the  polyhedrons  S-A  B  c,  s'-A'b'c',  if  the  lines  join- 
ing the  vertices  A  and  A',  B  and  J5',  etc.,  all  pass  through  the 
same  point  0,  and  are  bisected  in  that  point,  the  polyhedrons 
are  said  to  be  symmetrical  with  respect  to  0,  which  is  called 
their  center  of  symmetry. 

736.  Cor.  If  two  polyhedrons  are  symmetrical  with  respect 
to  a  center,  their  homologous  faces  are  severally  equal,  their 
dihedral  angles  are  severally  equal,  their  polyhedral  angles  are 
symmetrical,  and  the  polyhedrons  are  equivalent. 

737.  Definition.  A  polyhedron  A  B  cn-A'  is  said  to  be 
symmetrical  tvith  respect  to  a  center  O,  if  its  vertices,  taken 
two  and  two,  are  symmetrical  with 
regard  to  0;  i.e.,  if  A  A',  BB',  etc.,  are 
each  bisected  in  the  same  point  0. 

738.  Cor.  A  polyhedron,  in  order 
to  have  a  center  of  symmetry,  must  have 
an  even  number  of  edges,  must  have 
its  homologous  edges  equal  and  parallel, 
its  homologous  plane  angles  and  dihedral  angles  equal,  and  its 
homologous  polyliedral  angles  symmetrical. 


SYMMETRICAL  POLYHEDRONS. 


361 


II.   SYMMETRY  WITH  RESPECT  TO  AN  AXIS. 

739.    Definition.     Two  polyhedrons  are  said  to  be  sym- 
metrical ivith  respect  to  an  axis,  when  this  axis  is  an  axis  of 
symmetry  for  the  corresponding  vertices 
of  the  two  polyhedrons. 


740.  Definition.  A  polyhedron  is 
said  to  be  symmetrical  tuith  respect  to  an 
axis,  when  this  line  is  an  axis  of  sym- 
metry for  the  corresponding  vertices  of 
the  polyhedron,  taken  two  and  two. 

Thus  the  polyhedron  AB'-'E'  is  sym- 
metrical with  respect  to   the  axis  00 
when  this  axis  is  an  axis  of  symmetry  for  each  of  the 
pairs  of  vertices,  A  and  D,  B'  and  E',  etc. 


o 


D' 


III.    SYMMETRY  WITH  RESPECT  TO  A  PLANE. 

741.  Definition.  Two  points  are  said  to  be  symmetrical 
with  respect  to  a  plane,  when  the  plane  is  perpendicular  to, 
and  bisects,  the  straight  line  joining  the  two  points. 


Exercise  938.  Every  intercept  between  two  opposite  faces  of  a 
polyhedron  having  a  center  of  symmetry,  and  passing  through  that 
center,  is  bisected  there. 

939.  Every  prism  whose  bases  are  polygons  symmetrical  with 
respect  to  a  point,  has  a  center  of  symmetry. 

940.  Where  is  the  center  of  symmetry  of  a  parallelopiped  ? 

941.  A  right  prism  whose  bases  are  symmetrical  with  respect  to 
a  center,  has  an  axis  of  symmetry. 

942.  A  rectangular  parallelopiped  has  three  axes  of  symmetry. 

943.  How  many  axes  of  symmetry  has  a  cube  ? 

944.  A  regular  pyramid  having  an  even  number  of  lateral  faces, 
has  an  axis  of  symmetry. 


362  GEOME  TRY.  — A  PPENDIX. 

742.  Definition.  Two  polyhedrons  are  said  to  be  sym- 
metrical ivitli  respect  to  a  plane,  when  this  plane  is  a  plane  of 
symmetry  for  the  corresponding  vertices  of  the  polyliedrons, 
taken  two  and  two. 

743.  Cor.  hi  order  that  two  polyhedrons  may  he  symmetri- 
cal with  respect  to  a  plane,  their  homologous  faces  must  he 
severally  equal,  their  dihedral  angles  must  he  equal,  their 
polyhedral  angles  must  he  symmetrical,  and  the  ijolyhedrons 
must  he  equivalent. 

744.  Definition.  A  polyhedron  is  said  to  be  symmetri- 
cal with  respect  to  a  plane,  when  the  plane  divides  it  into  two 
polyhedrons  symmetrical  with  respect  to  that  plane. 


MAXIMA  AND  MINIMA. 

745.  Definition.  A  magnitude  is  said  to  be  a  maximum 
or  minimum  according  as  it  is  the  greatest  or  least  of  a  given 
class. 

Thus  the  diameter  of  a  circle  is  a  maximum  among  all 
inscribed  straight  lines;  and,  among  all  the  straight  lines 
drawn  from  a  given  point  to  a  given  straight  line,  the  per- 
pendicular is  the  minimum. 

746.  Definition.  Isoperimetric  figures  are  those  which 
have  equal  perimeters. 

Exercise  945.  How  many  planes  of  symmetry  has  any  right  prism  ? 

946.  How  many  has  a  parallelopiped  ? 

947.  Has  a  cylinder  a  plane  of  symmetry  ?     A  sphere  ? 

948.  Show  that,  if  a  polyhedron  has  two  planes  of  symmetry  that 
are  perpendicular  to  each  other,  their  common  intersection  is  an 
axis  of  symmetry. 

949.  Show  that,  if  a  polyhedron  has  three  planes  of  symmetry, 
the  point  common  to  the  three  planes  is  a  center  of  symmetry. 


MAXIMA    AND  MINIMA 


363 


Proposition  I.     Theorem. 

747.  Of  all  triangles  having  two  given  sides,  that 
in  which  tlxese  sides  are  perpendicular  to  each  other 
has  the  maximum  area. 


Given:  In  triangles  ABC,  A'BC,  AB  eqnal  to  A^B,  BC  equal 
to  B c,  but  only  A B  perpendicular  to  BC ; 

To  Prove :  Triangle  ABC  is  greater  tlian  triangle  a'b c. 

Draw  A' I)  ±  to  BC. 

Since  AB  =z  A'B,  (Hyp.) 

but  A'B  >  A'B,  (93) 

the  altitude  AB  ^  the  altitude  A'D; 

.:  Aabc>  AA'BC.  Q.E.D.     (333) 

Exercise  950.  Two  lines,  whose  lengths  are  a  and  b  respectively, 
being  given,  find  the  length  of  the  third  line  that  will  form  with  them 
the  maximum  triangle. 

951.  Of  all  triangles  of  given  base  and  area,  the  isosceles  has  the 
greatest  vertical  angle. 

952.  Of  all  triangles  of  given  base  and  vertical  angle,  the  isosceles 
is  the  greatest. 

953.  Of  all  triangles  of  given  altitude  and  vertical  angle,  the  isos- 
celes is  the  least. 

954.  Of  all  triangles  of  given  base  and  vertical  angle,  the  isosceles 
has  the  greatest  perimeter. 

955.  Divide  a  given  arc  into  two  parts  such  that  the  sum  of  the 
chords  subtending  them  shall  be  a  maximum. 


364  GEOMETRY.  — APPENDIX. 

Proposition   II.     Theorem. 

748.   Of  all  equivalent  triangles  on  the  same  base, 
the  isosceles  triangle  has  the  Tuinimum  perimeter. 


Given :  In  ^  J5  C,  A^B  C,  equivalent  triangles  on  the  base  B  C, 
AB  equal  to  AC ; 

To   Prove  :     AB -\-  AC -\-  BC  <  A^B  +  A'C-\-  BC. 

Produce  B A  ^o  that  AD  =  AB=  AC;  join  DC  and  AA\ 
and  produce  ^  J.'  to  ^. 

Since  ABC  and  A^BC  are   equivalent,  and  on  the  same 
base,  (Const.) 

ABC  and  A^B C  have  the  same  altitude,  (332) 

(since  otherwise  they  would  not  be  equivalent;) 

.-.   AA^El^  II  to  BC; 

.'.  EB  =  EC;  (147) 

.-.  ^^is±  to  DC;  (97) 

.-.   A^D=.A'C.  (96) 

But  BD  <  A'B-{-  A'D; 

i.e.,  AB  -}-AC<  A'B-{-  a'c. 

.'.   AB -{- AC -{- BC  <  A'B -\- A'C-\- BC.  Q.E.D. 

749.    CoR.    Of  all  equivalent  triangles,  that  which  is  equi- 
lateral has  the  least  perimeter. 


MAXIMA   AND  MINIMA.  365 

Proposition   III.     Theorem. 

750.   Of  all  isoperiinetric  triangles  on  the  same  base, 
that  which  is  isosceles  has  the  maximuTn  area. 


Given:  Two  isoperimetric  triangles  ABC,  A'BC,  and  AB  equal 
to  A  c; 

To  Proves  Triangle  ABCi^  greater  than  triangle  A^BC. 

Draw  AD  1.  to  BC,  and  A^E  II  to  BC,  and  meeting  AD  pro- 
duced if  necessary  in  E;  also  join  EB,  EC. 
Since  A  ^'^C,  z 5  (7,  have  the  same  altitude,     (Const.) 


A  A'BC=o=AEBC. 

(332) 

But  Z 5  (7  is  an  isosceles  A ; 

(96) 

.-.   A'B-\-A'C>EB-]-EC; 

(748) 

.'.  AB-^AC>EB,-\-EC, 

(Hyp.) 

(since  AB  -{-  AC=  a'b -f  A'c;) 

.'.  AB>EB; 

(Ax.  7) 

.:  AD>ED; 

(99) 

.'.  A  ABOAebcov  AA'BC. 

Q.E.D.       (333) 

751.  CoR.  Of  all  isoperiinetric  triangles,  that  which  is 
equilateral  has  the  maximum  area. 

Exercise  956.  Find  the  area  of  the  maximum  triangle  whose  perim- 
eter is  n  feet. 

957.  Of  all  triangles  inscribed  in  a  given  circle,  the  equilateral  has 
the  greatest  perimeter. 


366  GEOMETR  Y.  —  APPENDIX. 

Proposition   IV.     Theorem. 

752.  Of  all  polygons  formed  of  sides  all  given  hut 
one,  the  majciinum  can  be  inscribed  in  a  semicircle 
having  the  undetermined  side  as  diameter. 


Given:  ABCDEF,  the  maximum  polygon,  having  the  given 
sides  AB,  BC,  CD,  DE,  EF,  and  an  undetermined  side  AF ; 

To  Prove :  ABCDEF  can  be  inscribed  in  a  senricircle. 

Join  any  vertex,  as  C,  with  A  and  F.  Then  the  triangle 
A  CF  must  be  the  maximum  of  all  triangles  formed  with 
the  given  sides  A  G  and  CF. 

For  otherwise,  by  changing  /.  ACF,  leaving  AC  and  CF 
unchanged,  we  could  increase  A  A  CF,  leaving  the  rest  of 
the  polygon  unchanged. 

That  is,  the  whole  polygon  AB---  F  would  be  increased, 
which  it  cannot  be,  since  AB  •"  F  is  a  maximum. 

Hence  A  ACF  must  be   the   maximum   triangle  formed 
with  the  given  sides  A  C,  CF; 

.'.  Z  ACF  is  a  rt.  Z,  (747) 

(since  otherwise  A  ACF  would  not  be  a  maximum  ;) 

.-.  C  is  on  the  semicircumf.  whose  diam.  is  AF; 

.:  each  vertex  of  ABCDEF  is  on  that  semicircumf.    q.e.d. 

Exercise  958.  Of  all  parallelograms  of  a  given  base  and  area,  the 
rectangle  has  the  least  perimeter. 

959.  Of  all  rectangles  of  given  area,  the  square  has  the  least  perim- 
eter. 


MAXIMA   AND  MINIMA. 


367 


Proposition   Y.     Theorem. 

753.  Of  all  polygons  forfned  with  given  sides,  that 
which  can  he  inscribed  in  a  circle  is  the  maximum. 


Given:    Two   mutually  equilateral   polygons,   ABODE  and 
A'b'c'd'e',  of  which  only  ABODE  can  be  inscribed  in  a  circle; 

To  Prove :  ABODE  is  greater  than  A'b'o'd'e'. 

Draw  the  diameter  AF,  and  join  FO,  FD. 

Upon  0'd'{=  OD)  construct  Af'o'd'=  AFOD,  and  join^'i^'. 

Since  polygon  ABOF  is  inscribed  in  a  semicircle, 

ABOF>  A'B'O'F'; 

similarly  AEDF>  a'e'd'f' ; 

.'.   ABCFDE>  A'B'O'F'd'e';  (Ax.  4) 

.-.    ABOFDE-FOD>  A'b'O'F'D'e'-F'O'D' ;      (Ax.  5) 

Q.E.D. 


(752) 


.e.,   ABODE>  a'b'o'd'e'. 


754.  Scholium.  The  area  of  the  inscribed  polygon  will 
be  the  same  in  whatever  order  the  sides  are  arranged. 
For  these  sides  are  chords  which  cut  off  equal  segments, 
in  whatever  order  they  occur;  and  the  polygon  is  the 
difference  between  the  circle  and  these  segments. 

Exercise  960,  Of  all  rectangles  that  can  be  inscribed  in  a  given 
circle,  the  greatest  is  a  square. 


368  GEOMETRY.  — APPENDIX. 

Proposition  VI.     Theorem. 

755.  Of  isoperimetric  polygons  of  the  same  numher 
of  sides,  the  maxiTnum  is  a  regular  polygon. 


Given:  A  BCD,  the  maximum  of  isoperimetric  polygons  of  n 


To  Prove :       ABCD  is  si,  regular  polygon. 

The  polygon  ABCD  must  be  equilateral. 

For  if  any  two  of  its  sides,  BA,BC,  were  unequal,  then  upon 
^(7  as  base  we  could  construct  an  isosceles  triangle  B'AC, 
having  the  sum  of  its  sides,  B'A,  B'c,  equal  to  BA-\-BC. 
The  triangle  ^Uc  would  be  greater  than  BAC  (748),  and 
therefore  the  polygon  AB'CJ)  would  be  greater  than  the 
maximum  polygon  ABCD.  But  this  is  impossible;  hence 
ABCD  must  be  equilateral.  It  can  also  be  inscribed  in  a 
circle  (753)  ;  hence  it  is  a  regular  polygon  (380). 

Exercise  961.  Of  all  triangles  having  the  same  vertical  angle,  and 
whose  bases  pass  through  a  given  point,  that  whose  base  is  bisected 
by  the  given  point  is  least. 

962.  The  rectangle  contained  by  the  segments  of  a  line  is  a  maxi- 
mum when  the  segments  are  equal. 

963.  Through  a  given  point  within  a  given  circle,  draw  the  maxi- 
mum and  minimum  chords  that  pass  through  that  point. 

964.  From  a  given  point  without  a  given  circle,  draw  the  secant 
whose  outer  segment  is  a  minimum.     What  about  its  inner  segment  ? 


MAXIMA   AND  MINIMA.  369 

Proposition  VII.     Theorem. 

756.  Of  isoperimetric  regular  polygons,  that  having 
the  greatest  nufnher  of  sides  is  the  maximum. 


Given :  A  regular  polygon  ABC,  of  n  sides ; 

To  Prove :  ABC  is  less  than  an  isoperimetric  regular  polygon 
of  n  +  1  sides. 

In  one  of  the  sides,  as  AB,  take  any  point  P. 

We  may  now  regard  the  given  polygon  as  an  irregular 
polygon  of  n -4-1  sides,  in  which  the  sides  AP,  PB,  make 
with  each  other  an  angle  equal  to  two  right  angles. 

This  irregular  polygon  is  less  than  the  regular  polygon 
of  the  same  perimeter  and  having  n-{-l  sides  (755)  ;  that  is, 
a  regular  polygon  of  n  sides  is  less  than  the  isoperimetric 
regular  polygon  of  n  -{-1  sides,     q.e.d. 

757.  Cor.  The  circle  contains  a  maximum  area  ivithin  a 
given  perimeter. 

Exercise  9()o.  On  the  circumference  of  a  given  circle  find  the 
point  such  that  the  sum  of  the  squares  of  its  distances  from  two 
given  points  without  the  circle  shall  be  a  minimum. 

966.  Of  two  given  circles,  one  lies  wholly  within  the  other.  Find 
the  maximum  and  minimum  chords  of  the  outer  that  are  tangent  to 
the  inner  circle. 

967.  A  line  ABC  is  perpendicular  to  an  indefinite  line  CM.  Find 
the  point  P in  CiHf  at  which  the  angle  APE  is  a  maximum. 

968.  Find  a  point  within  a  quadrilateral  such  that  the  sum  of  the 
lines  drawn  from  that  point  to  the  vertices  shall  be  a  minimum. 


370 


GEOMETR  Y.  —  APPENDIX. 


Proposition  YIII.     Theorem. 

758.  Of  two  regular  -polygons  having  equal  areas, 
that  having  the  greater  nujnher  of  sides  has  the  less 
perimeter. 


Given :  P  and  Q,  regular  polygons  of  tlie  same  area,  but  Q  with 
the  greater  number  of  sides ; 

To  Prove :  Perimeter  of  Q  is  less  than  perimeter  of  P. 

Let  i?  be  a  regular  polygon  having  the  same  perimeter  as 
Q  and  the  same  number  of  sides  as  P. 

Then  since  Q>  E,  (756) 

but  Q  =  P,  (Hyp.) 

P>Ji; 

.'.  the  perimeter  ofR<  the  perimeter  of  P;  (346) 

.-.  the  perimeter  of  Q  <  the  perimeter  of  P.  q.e.d. 

759.    Cor.    TTie  circumference  of  a  circle  is  less  than  the 
perimeter  of  any  polygon  of  equal  area. 


Exercise  969.  Through  a  given  point  within  a  given  angle  draw 
the  intercept  that  cuts  off  the  triangle  of  maximum  area. 

970.  Through  a  point  of  intersection  of  two  circles  draw  that  inter- 
cept between  the  two  circumferences  which  is  a  maximum. 

971.  In  a  given  line  find  a  point  such  that  the  sum  of  its  distances 
from  two  given  points  without  the  line,  and  on  the  same  side  of  it, 
shall  be  a  minimum. 

972.  In  a  given  line  find  the  point  such  that  the  tangents  drawn 
from  it  to  a  given  circle  contain  the  maximum  angle. 


INDEX  OF  DEFINITIONS. 


PAGE 

PAGE 

Alternation, 

122 

Altitude  of  prism, 

262 

Angle, 

15 

of  pyramid. 

275 

acute, 

17 

Analysis, 

71, 105 

adjacent, 

16 

Antecedent, 

113 

alternate, 

51 

Apothem, 

199 

at  center  of  circle, 

128 

Arc, 

78 

of  regular  polygon. 

200 

major  and  minor, 

85 

complementary, 

17 

Area, 

165 

corresponding. 

51 

Axiom, 

19 

difference  of, 

16 

Axis  of  circle, 

301 

dihedral, 

240 

of  cone, 

297 

equal, 

16 

of  cylinder. 

294 

exterior, 

51 

of  symmetry. 

357 

face. 

251 

Base  of  triangle. 

28 

including. 

16 

of  polygon. 

165 

inscribed. 

131 

Bisector, 

38 

interior. 

61 

Center  of  polyhedrons, 

360 

oblique, 

17 

of  regular  polygon, 

199 

obtuse. 

17 

of  symmetry  of  plane  figures,  355 

of  the  line  and  plane, 

249 

Chord, 

78 

plane, 

15 

Circle, 

78 

polyhedral, 

251 

of  sphere. 

301 

right. 

17 

great  or  small, 

301 

spherical. 

309 

sector  of, 

210 

straight, 

16 

segment  of. 

210 

sum  of. 

16 

Circumference, 

15,  78 

supplementary. 

17 

Commensurable, 

112 

trihedral, 

251 

Common  measure. 

112 

vertical. 

16 

Complement, 

17 

Altitude  of  triangle. 

74 

CONCYCLIC, 

160 

of  cone. 

297 

Cone, 

297 

of  cylinder. 

294 

altitude  of. 

297,  33(> 

of  frustum  of  pyramid, 

275 

base  of. 

297 

of  polygon. 

165 

circular, 

297 

371 


372 


INDEX   OF  DEFINITIONS. 


PAGE 

PAGE 

Cone,  axis  of  circular, 

297 

Enunciation, 

20 

frustum  of, 

337 

Figure, 

13 

lateral  area  of, 

330 

equal, 

29 

lateral  surface  of, 

297 

equivalent, 

165 

of  revolution, 

297 

geometrical, 

13 

similar. 

337 

isoperimetric. 

362 

slant  height  of, 

336 

plane. 

13 

truncated. 

337 

Frustum  of  cone, 

337 

vertex  of, 

297 

altitude  of, 

337 

Conical  surface, 

297 

of  pyramid. 

275 

Consequent, 

113 

altitude  of, 

275 

Constant, 

117 

Generatrix,                        293, 297 

Construction, 

21 

Geometrical  concepts, 

12 

Corollary, 

18 

Geometry, 

12 

Cube, 

273 

of  space, 

225 

Curve, 

12 

of  three  dimensions, 

225 

Cylinder, 

294 

plane, 

13 

axis  of, 

294 

solid. 

225 

circular. 

294 

Hexahedron, 

261 

lateral  area  of, 

332 

Homologous, 

144 

oblique, 

294 

Hypotenuse, 

35 

of  revolution, 

294 

Hypothesis, 

20 

similar, 

333 

Icosahedron, 

261 

radius  of, 

294 

Incommensurable, 

112 

right, 

294 

Intercept, 

62 

Cylindrical  surface, 

293 

Inversion, 

122 

Demonstration, 

21 

Limit, 

117 

Diagonal, 

61 

Line, 

12 

Diagram, 

13 

broken. 

12 

Diameter, 

78 

curved. 

12 

Dihedral  angle, 

240 

of  centers. 

92 

equal. 

240 

parallel. 

51 

plane  angle  of, 

240 

parallel  to  plane. 

235 

right. 

^0 

perpendicular  to  plane. 

227 

Direction, 

14 

straight. 

12 

of  parallel  lines, 

57 

Locus, 

102 

Directrix, 

293,  297 

LUNE, 

324 

Distance,  central. 

92 

Magnitude, 

13 

equal, 

15 

equal, 

13 

of  point  from  line. 

47 

Maximum, 

362 

of  point  from  plane. 

230 

Median, 

74 

on  sphere, 

303 

Minimum, 

362 

polar. 

303,304 

Nappe,  upper  and  lower 

297 

Dodecahedron, 

261 

Numerical  measure. 

112 

Element  of  surface, 

293 

Octahedron, 

261 

INDEX    OF  DEFINITIONS, 


373 


Parallel  lines, 

PAGE 

51 

Proportion,  continued, 

PAGE 

124 

planes, 

235 

Proportional, 

114 

Parallelogram, 

61 

fourth, 

115 

Parallelopiped, 

2(!7 

inversely. 

152 

rectangular, 

267 

mean. 

119 

right, 

267 

third, 

119 

Perimeter, 

74 

Pyramid, 

275 

Plane, 

13 

altitude  of. 

275 

Point, 

12 

frustum  of. 

275 

Polar  distance, 

303 

regular, 

275 

triangle. 

313 

slant  height  of, 

275 

Pole, 

301 

triangular,  etc., 

275 

Polygon, 

59 

truncated. 

275 

center  of  regular, 

1<)9 

Quadrilateral, 

61 

circumscribed. 

101 

Quantity, 

111 

convex, 

59 

Radius  of  circle, 

15 

inscribed. 

101 

of  regular  polygon, 

199 

radius  of  regular. 

199 

of  sphere, 

300 

regular, 

198 

Ratio, 

113 

similar, 

144 

extreme  and  mean, 

158 

spherical. 

310 

inverse  or  reciprocal, 

152 

diagonal  of. 

311 

Rectangle, 

61 

stellate. 

215 

Rhombus, 

61 

Polyhedral  angle, 

251 

Scholium, 

19 

edge,  face,  A^ertex, 

251 

Secant, 

90 

equal  or  symmetrical, 

255 

Sector, 

210 

face  angle  of. 

251 

similar. 

210 

Polyhedron, 

201 

Segment,  harmonic. 

145 

convex, 

261 

of  circle. 

210 

equal,  equivalent, 

262 

of  a  line. 

140 

regular. 

286 

of  sphere, 

349 

similar. 

262 

similar, 

210 

volume  of, 

261 

Semicircle, 

80 

Postulate, 

19 

Sbmicircumferbnce, 

80 

Prism, 

262 

Solid, 

11 

altitude  of. 

262 

Space-concepts, 

12 

edge,  faces. 

262 

Sphere, 

300 

lateral  or  convex  surface, 

2()3 

tangent  to  plane, 

"307 

regular, 

262 

tangent  to  sphere. 

307 

right. 

262 

Spherical  angle. 

309 

right  section  of, 

262 

degree. 

325 

truncated, 

2(56 

excess. 

316 

Problem, 

18 

polygon. 

310 

Projection, 

179,  247 

pyramid, 

349 

Proportion, 

114 

sector. 

348 

374 


INDEX   OF  DEFINITIONS. 


PAGE 

PAGE 

Spherical  segment, 

349 

Triangle,  obtuse, 

29 

triangle, 

311 

right, 

29 

Square, 

61 

scalene, 

28 

Supplement, 

17 

spherical, 

311 

Surface, 

11 

birectangular, 

316 

Symmetry, 

355-362 

symmetrical, 

317 

Synthesis, 

71 

trirectangular, 

316 

Tangent, 

90 

supplemental, 

315 

circles. 

92 

Trihedral  angle, 

261 

spheres, 

307 

Ungula, 

324 

Tetrahedron, 

261 

Unit  of  area, 

168 

Theorem, 

18 

of  length, 

111 

converse, 

24 

of  volume. 

273 

Transversal, 

51 

Variable, 

117 

Trapezium, 

61 

Vertex, 

15 

Trapezoid, 

61 

Zone, 

346 

Triangle, 

28 

altitude  oi, 

346 

acute, 

29 

bases  of. 

346 

equilateral, 

28 

of  one  base, 

346 

isosceles. 

28 

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